NCERT Solutions for Class 11 Maths Chapter 10 - Conic Sections
Exercise 11.1
Directions (Q. Nos. 1 to 5): In each of the following find, the equation of the circle with:
1. Centre (0, 2) and radius 2.
Sol. Given centre = (h, k) = (0, 2) and radius r = 2
∴ Equation of circle by using (x – h)2 + (y –k)2 = r2 is
(x – 0)2 + (y – 2)2 = 22
⇒ x2 + y2 + 4 – 4y = 4
[∵ (a – b)2 = a2 + b2 – 2ab]
⇒ x2 + y2 – 4y = 0
∴ The equation of circle is x2 + y2 – 4y = 0.
2. Centre (– 2, 3) and radius 4.
Sol. Given centre = (h, k) = (– 2, 3) and radius r = 4
∴ Equation of circle by using (x – h)2 + (y – k)2 = r2 is
(x + 2)2 + (y – 3)2 = 42
⇒ x2 + 4 + 4x + y2 + 9 – 6y = 16
[∵ (a + b)2 = a2 + b2 + 2ab and (a – b)2
= a2 + b2 – 2ab]
⇒ x2 + y2 + 4x – 6y + 13 – 16 = 0
⇒ x2 + y2 + 4x – 6y – 3 = 0
∴ The equation of circle is
x2 + y2 + 4x – 6y – 3 = 0.
$$\textbf{3. Center}\bigg(\frac{\textbf{1}}{\textbf{2}},\frac{\textbf{1}}{\textbf{4}}\bigg)\space\textbf{and radius}\frac{\textbf{1}}{\textbf{12}}.\\\textbf{Sol.\space}\text{Given centre = (h, k) =}\bigg(\frac{1}{2},\frac{1}{4}\bigg)\\\text{and radius r}=\frac{1}{12}$$
∴ Equation of circle by using (x – h)2 + (y – k)2 = r2 is
$$\bigg(x-\frac{1}{2}\bigg)^2+\bigg(y-\frac{1}{4}\bigg)^2=\bigg(\frac{1}{12}\bigg)^{2}$$
$$\Rarr\space x^2+\frac{1}{4}-2.\frac{1}{2}x+y^2+\frac{1}{16}-2.\frac{y}{4}\\=\frac{1}{144}$$
[∵ (a – b)2 = a2 + b2 – 2ab]
$$\Rarr\space x^2+y^2-x-\frac{y}{2}+\frac{1}{4}+\\\frac{1}{16}-\frac{1}{144}=0\\\Rarr\space x^2+y^2-x-\frac{y}{2}+\\\frac{36+9-1}{144}=0\\\Rarr\space x^2+y^2-x-\frac{y}{2}+\frac{44}{144}=0\\\Rarr\space x^2+y^2-x-\frac{y}{2}+\frac{11}{36}=0$$
⇒ 36x2 + 36y2 – 36x – 18y + 11 = 0
∴ The equation of circle is
36x2 + 36y2 – 36x + 8y + 11 = 0.
$$\textbf{4. Centre (1, 1) and radius}\sqrt{\textbf{2}}.$$
Sol. Given centre = (h, k) = (1, 1)
$$\text{and radius\space=\space r=}\space\sqrt{2}$$
∴ Equation of circle by using (x –h)2 + (y – k)2 = r2 is
$$(x-1)^2+(y-1)^2=(\sqrt{2})^2$$
⇒ x2 + 1 – 2x + y2 + 1 – 2y = 2
⇒ x2 + y2 – 2x – 2y = 0
∴ The equation of circle is
x2 + y2 – 2x – 2y = 0.
$$\textbf{5. Centre (– a, – b) and radius}\sqrt{\textbf{a}^\textbf{2}-\textbf{b}^\textbf{2}}\textbf{.}$$
Sol. Given centre = (h, k) = (– a, – b) and radius
$$r=\sqrt{a^2-b^2}$$
∴ Equation of circle by using (x – h)2 + (y – k)2 = r2 is
$$(x+a)^2+(y+b)^2=(\sqrt{a^2-b^2})^2$
⇒ x2 + a2 + 2ax + y2 + b2 + 2by = a2 – b2
⇒ x2 + y2 + 2ax + 2by + 2b2 = 0
∴ The equation of circle is
x2 + y2 + 2ax + 2by + 2b2 = 0.
Direction (Q. Nos. 6-9): In each of the following, find the centre and radius of the circle:
6. (x + 5)2 + (y – 3)2 = 36.
Sol. Given equation of circle is
(x + 5)2 + (y – 3)2 = 36 = (6)2
Comparing it with
(x – h)2 + (y – k)2 = r2
We get h = – 5, k = 3 and r = 6
Hence, centre = (h, k) = (– 5, 3) and radius r = 6.
7. x2 + y2 – 4x – 8y – 45 = 0.
Sol. Given equation of circle is
x2 + y2 – 4x – 8y – 45 = 0
Comparing with
x2 + y2 + 2gx + 2fy + c = 0
We get 2g = – 4, 2f = – 8, c = – 45
⇒ g = – 2, f = – 4, c = – 45
∵ As centre and radius of circle
x2 + y2 + 2gx + 2fy + c = 0 is
$$\text{C(– g, – f) and r =}\sqrt{g^2+f^2-c}$$
∴ Centre = (– g, – f) = (2, 4)
$$\text{and radius}=\sqrt{g^2+f^2-c}\\=\sqrt{(2)^2+(4)^2-(-45)}\\=\sqrt{4+1 6+45}=\sqrt{65}$$
8. x2 + y2 – 8x + 10y – 12 = 0.
Sol. Given equation of circle is
x2 + y2 – 8x + 10y – 12 = 0
Comparing with
x2 + y2 + 2gx + 2fy + c = 0
We get, 2g = – 8, 2f = 10, c = – 12
⇒ g = – 4, f = 5, c = – 12
∴ Centre = (– g, – f) = (4, – 5)
$$\text{and radius} = \sqrt{g^2+f^2-c}\\=\sqrt{(4)^2+(-5)^2-(-12)}\\=\sqrt{16+25+12}\\=\sqrt{53}$$
9. 2x2 + 2y2 – x = 0.
Sol. Given equation of circle is
2x2 + 2y2 – x = 0
On dividing by 2, we get
$$x^2+y^2-\frac{x}{2}=0$$
On comparing with x2 + y2 + 2gx + 2fy + c = 0, we get
$$2g=-\frac{1}{2},\text{2f = 0, c = 0}\\\Rarr\space \text{g =}-\frac{1}{4}, f=0, c=0$$
∴ Centre = (– g, – f) = (1/4, 0)
$$\text{and radius}=\sqrt{g^2+f^2-c}\\=\sqrt{\bigg(\frac{1}{4}\bigg)^2+0-0}\\=\sqrt{\frac{1}{16}}=\frac{1}{4}.$$
10. Find the equation of the circle, passing through (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.
Sol. Let equation of circle is
x2 + y2 + 2gx + 2fy + c = 0 …(i)
Equation (i) passes through the points (4, 1) and (6, 5) i.e., they will satisfy it. When point (4, 1)
(4)2 + (1)2 + 2g(4) + 2f(1) + c = 0
⇒ 17 + 8g + 2f + c = 0
⇒ 8g + 2f + c + 17 = 0 …(ii)
When point (6, 5),
(6)2 + (5)2 + 2g(6) + 2f(5) + c = 0
⇒ 36 + 25 + 12g + 10f + c = 0
⇒ 12g + 10f + c + 61 = 0 …(iii)
As centre (– g, – f) of equation (i) lies on the line
4x + y = 16
∴ 4(– g) + (– f) = 16
⇒ 4g + f + 16 = 0 …(iv)
Now, subtract equation (ii) from equation (iii), we get
4g + 8f + 44 = 0 …(v)
Now, subtract equation (iv) from equation (v),
we get
7f + 28 = 0
⇒ 7f = – 28
⇒ f = – 4
From equation (iv),
4g – 4 + 16 = 0
⇒ 4g + 12 = 0
⇒ 4g = – 12
⇒ g = – 3
From equation (ii),
8(– 3) + 2(– 4) + c + 17 = 0
⇒ – 24 – 8 + c + 17 = 0
⇒ c – 32 + 17 = 0
⇒ c – 15 = 0
⇒ c = 15
On putting the values of g, f and c in equation (i), we get required equation of circle
x2 + y2 + 2(– 3)x + 2(– 4)y + 15 = 0
⇒ x2 + y2 – 6x – 8y + 15 = 0.
11. Find the equation of the circle passing through the points (2, 3) and (– 1, 1) and whose centre is on the line x – 3y – 11 = 0.
Sol. Let equation of circle is
x2 + y2 + 2gx + 2fy + c = 0 …(i)
Equation (i) passes through the points (2, 3) and (– 1, 1) i.e., they will satisfy it.
When point (2, 3),
(2)2 + (3)2 + 2g(2) + 2f(3) + c = 0
⇒ 4g + 6f + c + 13 = 0 …(ii)
and when point (– 1, 1),
(– 1)2 + (1)2 + 2g(– 1) + 2f(1) + c = 0
⇒ – 2g + 2f + c + 2 = 0 …(iii)
As centre (– g, – f) lies on the line
x – 3y – 11 = 0
i.e., (– g) – 3(– f) – 11 = 0
⇒ – g + 3f – 11 = 0 …(iv)
Now, subtract equation (iii) from equation (ii),
we get
6g + 4f + 11 = 0 …(v)
From equation (iv), g = 3f – 11 put in equation (v), we get
6(3f –11) + 4f + 11 = 0
⇒ 18f – 66 + 4f + 11 = 0
⇒ 22f – 55 = 0
$$\Rarr\space f=\frac{55}{22}\\\Rarr\space f=\frac{5}{2}\\\text{From equation (iv),}\\g=3×\frac{5}{2}-11\\=\frac{15-22}{2}=\frac{-7}{2}\\\text{From equation (ii),}\\4×\bigg(-\frac{7}{2}\bigg)+6\bigg(\frac{5}{2}\bigg)+c+13=0$$
⇒ – 14 + 15 + c + 13 = 0
⇒ c + 14 = 0
⇒ c = – 14
Put the values of g, f and c in equation (i) to get the required equation of circle
$$x^2+y^2+2x\bigg(-\frac{7}{2}\bigg)+\\2y\bigg(\frac{5}{2}\bigg)-14=0$$
⇒ x2 + y2 – 7x + 5y – 14 = 0
∴ The equation of circle is
x2 + y2 – 7x + 5y – 14 = 0.
12. Find the equation of the circle with radius 5 whose centre lies on X-axis and passing through the point (2, 3).
Sol. Let centre of circle be (h, 0), as centre of circle is on X-axis, so y-coordinate will be zero.
The distance between any point on the circle to the centre = Radius
$$\therefore\space\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=r\\\therefore\space\sqrt{(h-2)^2+(0-3)^2}=5$$
(∵ x1 = 2, y1 = 3, x2 = h, y2 = 0)
⇒ h2 + 4 – 4h + 9 = 25
(Squaring both sides)
⇒ h2 – 4h + 13 – 25 = 0
⇒ h2 – 4h – 12 = 0
⇒ h2 – 6h + 2h – 12 = 0
⇒ h(h – 6) + 2(h – 6) = 0
⇒ (h – 6) (h + 2) = 0
⇒ h – 6 = 0 or h + 2 = 0
⇒ h = 6 or – 2
∴ Centre is (6, 0) or (– 2, 0).
Hence, equation of circle by using
(x – h)2 + (y – k)2 = r2
when (h, k) = (6, 0) and r = 5,
(x – 6)2 + (y – 0)2 = 52
⇒ x2 + 36 – 12x + y2 = 25
⇒ x2 + y2 – 12x + 11 = 0
And when (h, k) = (– 2, 0) and r = 5
⇒ (x + 2)2 + (y – 0)2 = 52
(∵ (x – h)2 + (y – k)2 = r2)
⇒ x2 + 4 + 4x + y2 = 25
⇒ x2 + y2 + 4x – 21 = 0
∴ The equation of the circle is
x2 + y2 – 12x + 11 = 0
and x2 + y2 + 4x – 21 = 0.
13. Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.
Sol. Since, the circle passes through the points
O(0, 0), A(a, 0) and B(0, b).
Let general equation of circle is
x2 + y2 + 2gx + 2fy + c = 0 …(i)
Equation (i) passes through the points O(0, 0), A(a, 0) and B(0, b) i.e., they will satisfy it.
When point (0, 0),
0 + 0 + 0 + 0 + c = 0
⇒ c = 0
When point (a, 0),
a2 + 0 + 2ag + 0 + 0 = 0
⇒ a2 + 2ag = 0
$$\Rarr\space g=-\frac{a}{2}$$
and when point (0, b),
0 + b2 + 0 + 2bf + 0 = 0
⇒ b2 + 2bf = 0
$$\Rarr\space f=-\frac{b}{2}$$
Put the value of g, f and c in equation (i) to get the required equation of circle.
$$x^2+y^2+2x\bigg(-\frac{a}{2}\bigg)+\\2y\bigg(-\frac{b}{2}\bigg)+0=0$$
⇒ x2 + y2 – ax – by = 0
∴ The equation of the circle is
x2 + y2 – ax – by = 0.
14. Find the equation of a circle with centre (2, 2) and passing through (4, 5).
Sol. Here, centre of the circle (h, k) = (2, 2) and point on the circle P(x, y) = (4, 5)
$$r=\sqrt{(2-4)^2+(2-5)^2}\\=\sqrt{4+9}=\sqrt{13}\\\bigg[\because r=\sqrt{(x-h)^2+(y-k)^2}\bigg]$$
∴ Equation of circle is
$$\therefore\space(x-2)^2+(y-2)^2=(\sqrt{13})^2\\\text{[Given (h,k)=(2,2) and r}=\sqrt{13}]$$
⇒ x2 + 4 – 4x + y2 + 4 – 4y = 13
⇒ x2 + y2 – 4x – 4y – 5 = 0
Hence, the equation of circle is
x2 + y2 – 4x – 4y – 5 = 0.
15. Does the point (– 2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?
Sol. Given, equation of circle is
x2 + y2 = 25
Let S = x2 + y2 – 25 …(i)
Put (x, y) = (– 2.5, 3.5) in equation (i),
⇒ S = (– 2.5)2 + (3.5)2 – 25
= 6.25 + 12.25 – 25
= 18.50 – 25
= – 6.50 < 0
∴ Point lies inside the circle.
Exercise 11.2
Direction (Q. Nos. 1 to 6): In each of th following, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum:
1. y2 = 12x
Sol. Given, equation of parabola is
y2 = 12x
which is of the form
y2 = 4ax
i.e., focus lies on the positive direction of X-axis.
Here 4a = 12
⇒ a = 3
∴ The co-ordinate of focus = (a, 0) = (3, 0)
Axis = X-axis
∴ The equation of directrix,
x = – a
⇒ x = – 3
Length of latus rectum = 4a = 4 × 3 = 12.
2. x2 = 6y
Sol. Given equation of parabola is x2 = 6y which is of the form x2 = 4ay i.e., focus lies on the positive direction of Y-axis.
Here, 4a = 6
$$\Rarr\space a=\frac{6}{4}=\frac{3}{2}\\\therefore\text{ The co-ordinate of focus = (0, a) =}\\\bigg(0,\frac{3}{2}\bigg)$$
Axis = Y-axis
Thus, the equation of directrix,
y = – a
$$\Rarr\space y=-\frac{3}{2}\\\text{Length of latusrectum = 4a =}\\4×\frac{3}{2}=6$$
3. y2 = – 8x
Sol. Given equation of parabola is y2 = – 8x which is of the form y2 = – 4ax i.e., focus lies on the negative direction of X-axis.
Here, – 4a = – 8
⇒ a = 2
∴ Coordinate of focus = (– a, 0) = (– 2, 0)
Axis = X-axis
Equation of directrix,
x = a
⇒ x = 2
Length of latus rectum = 4a = 4 × 2 = 8.
4. x2 = – 16y
Sol. Given equation of parabola is x2 = – 16y, which is of the form x2 = – 4ay, i.e., focus lies on the negative direction of Y-axis.
Here, – 4a = – 16
⇒ a = 4
∴ Coordinate of focus = (0, – a) = (0, – 4)
Axis = Y-axis
∴ The equation of directrix,
y = a
⇒ y = 4
Length of latusrectum = 4a = 4 × 4 = 16.
5. y2 = 10x
Sol. Given equation of parabola is y2 = 10x, which is of the form y2 = 4ax i.e., focus lies on the positive direction of X-axis.
Here, 4a = 10
$$\Rarr\space a=\frac{5}{2}\\\therefore\space\text{Coordinate of focus = (a, 0)}\\ =\bigg(\frac{5}{2},0\bigg)$$
Axis = X-axis
Thus, equation of directrix,
x = – a
$$\Rarr\space x=-\frac{5}{2}\\\text{Length of latusrectum = 4a =}\\4×\frac{5}{2}=10.$$
6. x2 = – 9y
Sol. Given equation of parabola is x2 = – 9y, which is of the form x2 = – 4ay i.e., focus lies on the negative direction of Y-axis.
Here, – 4a = – 9
$$\Rarr\space a=\frac{9}{4}\\\therefore\space\text{Coordinate of focus}=(0,-a)\\=\bigg(0,-\frac{9}{4}\bigg)$$
Axis = Y-axis
Thus, equation of directrix,
y = a
$$\Rarr\space y=\frac{9}{4}$$
Length of latus thin rectum = 4a = 9.
Direction (Q. Nos. 7 to 12): In each of the following find the equation of the parabola that satisfies the given conditions.
7. Focus (6, 0) and directrix x = – 6.
Sol. Given, focus (6, 0) and directrix x = – 6
So, focus lies on the positive direction of X-axis i.e., equation of parabola will be of the form y2 = 4ax. Here a = 6.
Hence, the equation of the parabola is
y2 = 4 × 6x
⇒ y2 = 24x.
8. Focus (0, – 3) and directrix y = 3.
Sol. Given, focus = (0, – 3) and directrix y = 3
Since, focus lies on the negative direction of Y-axis i.e., equation of parabola will be of the form x2 = – 4ay. Here a = 3.
Hence, the equation of the parabola is
x2 = – 4(3)y
⇒ x2 = – 12y.
9. Vertex (0, 0) and focus (3, 0).
Sol. Given, vertex = (0, 0)
Focus = (3, 0)
Since, vertex is (0, 0) and focus lies on positive direction of X-axis. Hence, equation of parabola will be of the form y2 = 4ax. Here a = 3.
Hence, the equation of the parabola is
y2 = 4 × 3x
⇒ y2 = 12x.
10. Vertex (0, 0) and focus (– 2, 0).
Sol. Given, vertex = (0, 0)
Focus = (– 2, 0)
Since, vertex is (0, 0) and focus lies on the negative direction of X-axis. Hence, equation of parabola will be of the form
y2 = – 4ax with a = 2
Hence, the equation of the parabola is
y2 = – 4(2)x = – 8x.
11. Vertex (0, 0) passing through (2, 3) and axis is along X-axis.
Sol. Given, vertex = (0, 0)
Point = (2, 3)
Axis = X-axis
Since, point (2, 3) lies in first qudrant and axis is X-axis. hence, equation of parabola will be of the form y2 = 4ax, which passes through (2, 3) i.e.,
Put x = 2, y = 3 in y2 = 4ax
∴ (3)2 = 4a × (2)
$$\Rarr\space a=\frac{9}{8}$$
Hence, the equation of the parabola is
$$y^2=4\bigg(\frac{9}{8}\bigg)x\\\Rarr\space y^2=\frac{9}{2}x$$
12. Vertex (0, 0) passing through (5, 2) and symmetric with respect to Y-axis.
Sol. Given, vertex = (0, 0)
Point = (5, 2)
Since, point (5, 2) lies in first quadrant and axis is Y-axis. Therefore, parabola is symmetric with respect to Y-axis. Hence, equation of parabola will be of the form x2 = 4ay, which passes through
(5, 2) i.e.,
Put x = 5, y = 2 in x2 = 4ay
∴ (5)2 = 4a × 2
$$\Rarr\space a=\frac{25}{8}$$
Hence, the equation of the parabola is
$$x^2=4×\frac{25}{8}\\\Rarr\space x^2=\frac{25}{2}y.$$
Exercise 11.3
Question (Q. Nos. 1 to 9): In each of the question find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latusrectum of the ellipse.
$$\textbf{1.}\space\frac{\textbf{x}^\textbf{2}}{\textbf{36}}+\frac{\textbf{y}^\textbf{2}}{\textbf{16}}\textbf{= 1.}$$
$$\textbf{Sol.}\space\text{Given,}\space\frac{x^2}{36}+\frac{y^2}{36}=1$$
Here, the Denominator of is larger than the denominator of
$$\frac{\text{y}^\text{2}}{\text{16}}.$$
So, the major axis is along X-axis and minor axis is along Y-axis. On comparing the given equation with
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\space\text{we get}$$
a2 = 36 and b2 = 16
⇒ a = 6 and b = 4
Now, c2 = a2 – b2 = 36 – 16 = 20
$$\Rarr\space c=2\sqrt{5}$$
Here, major axis is along X-axis.
$$\therefore\space\text{Foci}=(\pm c,0)=(\pm2\sqrt{5},0)\\\text{Vertices}=(\pm a,0)=(\pm6,0)$$
Length of major axis = 2a = 2 × 6 = 12
Length of minor axis = 2b = 2 × 4 = 8
$$\text{Eccentricity, e}=\frac{c}{a}\\=\frac{2\sqrt{5}}{6}=\frac{\sqrt{5}}{3}\\\text{The length of latusrectum =}\frac{2b^2}{a}\\=\frac{2×16}{6}=\frac{16}{3}.$$
$$\textbf{2.}\space\frac{\textbf{x}^\textbf{2}}{\textbf{4}}+\frac{\textbf{y}^\textbf{2}}{\textbf{25}}\textbf{ = 1.}$$
Sol. Given equation of ellipse is
$$\frac{x^2}{4}+\frac{y^2}{25}=1\\\text{Here, the denominator of}\space\frac{x^2}{4}\\\text{is smaller than the denominator of}\\\frac{y^2}{25}.$$
So, the major axis is along Y-axis and minor axis is along X-axis.
On comparing the given equation with
$$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1,\space\text{we get}$$
b2 = 4 and a2 = 25
⇒ b =2 and a = 5
∵ c2 = a2 – b2 = 25 – 4 = 21
$$\Rarr\space c=\sqrt{21}$$
Here, major axis is along Y-axis. So,
$$\therefore\space\text{Foci = (0, ± c) =}(0,\pm\sqrt{21})$$
Vertices = (0, ± a) = (0, ± 5)
Length of major axis = 2a = 2 × 5 = 10
Length of minor axis = 2b = 2 × 2 = 4
$$\text{Eccentricity, e =}\frac{c}{a}=\frac{\sqrt{21}}{5}\\\text{The length of latusrectum =}\frac{2b^2}{a}\\=\frac{2×4}{5}=\frac{8}{5}.$$
$$\textbf{3.}\space\frac{\textbf{x}^\textbf{2}}{\textbf{16}}+\frac{\textbf{y}^\textbf{2}}{\textbf{9}}\textbf{=1.}$$
Sol. Given equation of ellipse is
$$\frac{x^2}{16}+\frac{y^2}{9}=1\\\text{Here, the denominator of is}\space\frac{x^2}{16}\\\text{greater than the denominator of}\\\frac{y^2}{9}.$$
So, the major axis is along X-axis and minor axis is along Y-axis.
On comparing given equation with
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$$
we get
a2 = 16 and b2 = 9
⇒ a = 4 and b = 3
Now,
c2 = a2 – b2 = 16 – 9 = 7
$$\Rarr\space c=\sqrt{7}$$
Here, major axis is along X-axis. So,
$$\therefore\space \text{Foci = (± c, 0) =}(\pm\sqrt{7},0)$$
Vertices = (± a, 0) = (± 4, 0)
Length of major axis = 2a = 2 × 4 = 8
Length of minor axis = 2b = 2 × 3 = 6
$$\text{Eccentricity, e =}\frac{c}{a}=\frac{\sqrt{7}}{4}\\\text{The length of latus rectum =}\\\frac{2b^2}{a}=\frac{2×9}{4}=\frac{9}{2}.$$
$$\textbf{4.}\space\frac{\textbf{x}^\textbf{2}}{\textbf{25}}+\frac{\textbf{y}^\textbf{2}}{\textbf{25}}\space\textbf{=\space1.}$$
Sol. Given equation of ellipse is
$$\frac{x^2}{25}+\frac{y^2}{100}=1\\\text{Here the denominator of}\space\frac{x^2}{25}\\\text{is smaller than the denominator of}\space\frac{y^2}{100}.$$
∴ The major axis is along Y-axis.
$$\text{On comparing given equation with}\\\space\frac{x^2}{b^2}+\frac{y^2}{a^2}=1.$$
we get
b2 = 25 and a2 = 100
⇒ b = 5 and a = 10
Now,
c2 = a2 – b2 = 100 – 25 = 75 = 25 × 3
$$\Rarr\space c=5\sqrt{3}$$
Here, major axis is along X-axis. So,
$$\therefore\space\text{Foci = (0, ± c) =}(0,\pm \space 5\sqrt{3})$$
Vertices = (0, ± a) = (0, ± 10)
The length of major axis = 2a = 2 × 10 = 20
The length of minor axis = 2b = 2 × 5 = 10
$$\text{Eccentricity e}=\frac{c}{a}=\\\frac{5\sqrt{3}}{10}=\frac{\sqrt{3}}{2}\\\text{The length of latus rectum =}\\\frac{2b^2}{a}=\frac{2×25}{10}=5.$$
$$\textbf{5.\space}\frac{\textbf{x}^\textbf{2}}{\textbf{49}}+\frac{\textbf{y}^\textbf{2}}{\textbf{36}}\space\textbf{= 1.}$$
Sol. Given, equation of ellipse is
$$\frac{x^2}{49}+\frac{y^2}{36}=1\\\text{Here,the denominator of}\space\frac{x^2}{49}\\\text{is larger than the denominator of}\space\frac{y^2}{36}.$$
So, the major axis is along X-axis.
On comparing given equation with
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$$
we get
a2 = 49, b2 = 36
⇒ a = 7, b = 6
Now, c2 = a2 – b2 = 49 – 36 = 13
$$\Rarr\space c=\sqrt{13}$$
Here major axis is along Y-axis. So,
$$\therefore\space\text{Foci}=(\pm c,0)=(\pm\sqrt{13},0)$$
Vertices = (± a, 0) = (± 7, 0)
The length of major axis = 2a = 2 × 7 = 14
The length of minor axis = 2b = 2 × 6 = 12
$$\text{Eccentricity = e}=\frac{c}{a}=\frac{\sqrt{13}}{7}\\\text{The length of latus rectum =}\space\frac{2b^2}{a}\\=\frac{2×36}{7}=\frac{72}{7}.$$
$$\textbf{6.\space}\frac{\textbf{x}^\textbf{2}}{\textbf{100}}+\frac{\textbf{y}^\textbf{2}}{\textbf{400}}\textbf{ = 1.}$$
Sol. Given equation of ellipse is
$$\frac{x^2}{100}+\frac{y^2}{400}=1\\\text{Here, the denominator of}\space\frac{x^2}{100}\\\text{is smaller than the denominator of}\space\frac{y^2}{400}.$$
So, the major axis is along Y-axis.
On comparing given equation with
$$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1,$$
we get
b2 = 100 and a2 = 400
⇒ b = 10 and a = 20
Now, c2 = a2 – b2
= 400 – 100 = 300
$$\Rarr\space c=10\sqrt{3}$$
Here, major axis is along Y-axis.
$$\therefore\space\text{Foci = (0, ± c) =}(0,\pm 10\sqrt{3})$$
Vertices = (0, ± a) = (0, ± 20)
the length of major axis = 2a = 2 × 20 = 40
the length of minor axis = 2b = 2 × 10 = 20
$$\text{Eccentricity, e}=\frac{c}{a}\\=\frac{10\sqrt{3}}{10}=\frac{\sqrt{3}}{2}\\\text{The length of latus rectum =}\frac{2b^2}{a}\\=\frac{2×100}{20}=10.$$
7. 36x2 + 4y2 = 144.
Sol. Given, equation of ellipse is
36x2 + 4y2 = 144
$$\Rarr\space\frac{36x^2}{144}+\frac{4y^2}{144}=1\\\text{(Divide by 144)}\\\Rarr\space\frac{x^2}{4}+\frac{y^2}{36}=1\\\text{Here,the denominator of}\space\frac{x^2}{4}\\\text{is smaller than the denominator of}\space \frac{y^2}{36}.$$
So, the major axis is along Y-axis.
On comparing given equation with
$$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1,$$
we get
b2 = 4 and a2 = 36
⇒ b = 2 and a = 6
and c2 = a2 – b2 = 36 – 4 = 32
$$\Rarr\space c= 4\sqrt{2}$$
Here, major axis is along Y-axis.
$$\therefore\space\text{Foci}=(0,\pm c)=(0,\pm4\sqrt{2})$$
Vertices = (0, ± a) = (0, ± 6)
The length of major axis = 2a = 2 × 6 = 12
The length of minor axis = 2b = 2 × 2 = 4
$$\text{Eccentricity, e =}\frac{c}{a}\\=\frac{4\sqrt{2}}{6}=\frac{2\sqrt{2}}{3}\\\text{The length of latus rectum =}\frac{2b^2}{a}\\=\frac{2×4}{6}=\frac{4}{3}.$$
8. 16x2 + y2 = 16.
Sol. Given, equation of ellipse is
16x2 + y2 = 16
$$\Rarr\space\frac{16x^2}{16}+\frac{y^2}{16}=1\\\text{(Divide by 16)}\\\Rarr\space\frac{x^2}{1}+\frac{y^2}{16}=1\\\text{Here, the denominator of}\space\frac{x^2}{1}\\\text{is smaller than the denominator of}\space\frac{y^2}{16}.$$
So, the major axis is along Y-axis.
On comparing given equation with
$$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1,$$
we get
b2 = 1 and a2 = 16
⇒ b = 1 and a = 4
Now, c2 = a2 – b2 = 16 – 1 = 15
$$\Rarr\space c=\sqrt{15}$$
Here, major axis is along Y-axis.
$$\text{Foci = (0, ± c) =}(0,\pm\sqrt{15})$$
Vertices = (0, ± a) = (0, ± 4)
The length of major axis = 2a = 2 × 4 = 8
The length of minor axis = 2b = 2 × 1= 2
$$\text{Eccentricity, e =}\frac{c}{a}=\frac{\sqrt{15}}{4}\\\text{The length of latus rectum =}\frac{2b^2}{a}\\=\frac{2×1}{4}=\frac{1}{2}.$$
9. 4x2 + 9y2 = 36.
Sol. Given, equation of ellipse is
4x2 + 9y2 = 36
$$\Rarr\space \frac{4x^2}{36}+\frac{9y^2}{36}=1\\\text{(Dividing by 36)}\\\Rarr\space \frac{x^2}{9}+\frac{y^2}{4}=1\\\text{Here, the denominator of}\space\frac{x^2}{9}\\\text{is greater than denominator of}\space\frac{y^2}{4}.$$
So, the major axis is along X-axis. On comparing given equation with
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\space\text{we get}$$
a2 = 9 and b2 = 4
⇒ a = 3 and b = 2
Now, c2 = a2 – b2 = 9 – 4 = 5
$$\Rarr\space c=\sqrt{5}$$
Here, major axis is along X-axis.
$$\text{Foci = (± c, 0) =}(\pm\sqrt{5},0)$$
Vertices = (± a, 0) = (± 3, 0)
The length of major axis = 2a = 2 × 3 = 6
The length of minor axis = 2b = 2 × 2 = 4
$$\text{Eccentricity, e =}\frac{c}{a}=\frac{\sqrt{5}}{3}\\\text{The length of latus rectum =}\frac{2b^2}{a}\\=\frac{2×4}{3}=\frac{8}{3}.$$
Direction (Q. Nos. 10 to 20): In each of the following questions find the equation for the ellipse that satisfies the given conditions:
10. Vertices (± 5, 0) and foci (± 4, 0).
Sol. Since, vertices (± 5, 0) and foci (± 4, 0) are on X-axis as y-coordinate of both points is zero.
Hence, equation of ellipse will be of the form
$$\frac{x}{a^2}+\frac{y}{b^2}=1\space\text{…(i)}$$
where, a = 5 and c = 4
(take a positve sign)
Now, c2 = a2 – b2
∴ (4)2 = (5)2 – b2
⇒ 16 = 25 – b2
⇒ b2 = 25 – 16
⇒ b2 = 9
Put the values of a2 = 25 and b2 = 9 in equation (i), we get
Equation of ellipse is
$$\frac{x^2}{25}+\frac{y^2}{9}=1.$$
11. Vertices (0, ± 13) foci (0, ± 5).
Sol. Since, vertices (0, ± 13) and foci (0, ± 5) are on Y-axis, x-coordinate of both points is zero.
Hence, equation of ellipse will be of the form
$$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\space\text{...(i)}$$
where, a = 13 and c = 5 (take positive sign)
∵ c2 = a2 – b2
⇒ (5)2 = (13)2 – b2
⇒ 25 = 169 – b2
⇒ b2 = 169 – 25
⇒ b2 = 144
Put the value of a2 = 169 and b2 = 144 in equation (i), we get
Equation of ellipse is
$$\frac{x^2}{144}+\frac{y^2}{169}=1.$$
12. Vertices (± 6, 0) and foci (± 4, 0).
Sol. Since, vertices (± 6, 0) and foci (± 4, 0) are on X-axis as y-coordinate of both points is zero.
Hence, equation of ellipse will be of the form
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\space\text{...(i)}$$
where, a = 6 and c = 4 (take positive sign)
∵ c2 = a2 – b2
⇒ (4)2 = (6)2 – b2
⇒ 16 = 36 – b2
⇒ b2 = 36 – 16 = 20
Put the values of a2 = 36 and b2 = 20 in equation (i), we get
Equation of ellipse is
$$\frac{x^2}{36}+\frac{y^2}{20}=1.$$
13. Ends of major axis (± 3, 0) and ends of minor axis (0, ± 2).
Sol. Since, major axis (± 3, 0) is along X-axis and minor axis (0, ± 2) is along Y-axis. Hence, equation of ellipse will be of the form
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\space\text{...(i)}$$
where, a = 3 and b = 2
(take positive sign)
Hence, equation (i) becomes
$$\frac{x^2}{3^2}+\frac{y^2}{2^2}=1\\\text{Equation of ellipse is}\\\frac{x^2}{9}+\frac{y^2}{4}=1$$
$$\textbf{14. Ends of major axis}\space(\textbf{0}, \pm\sqrt{\textbf{5}}),\\\textbf{ends of minor axis (± 1, 0).}$$
Sol. Since, major axis is along Y-axis and minor axis (± 1, 0) is along X-axis. Hence, equation of ellipse will be of the form
$$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\space\text{...(i)}\\\text{where a =}\sqrt{5}\space\text{and b = 1}\\\text{Hence, equation (i) becomes}\\\frac{x^2}{1^2}+\frac{y^2}{(\sqrt{5})^2}=1$$
Equation of ellipse is
$$\text{or}\space \frac{x^2}{1}+\frac{y^2}{5}=1.$$
15. Length of major axis 26, foci (± 5, 0).
Sol. Since, foci (± 5, 0) lies on X-axis, as y-coordinate is zero. Hence, equation of ellipse will be of the form
$$\frac{x^2}{a^2}+\frac{x^2}{b^2}=1\space\text{...(i)}$$
Given that length of major axis,
2a = 26
⇒ a = 13 and c = 5
∵ c2 = a2 – b2
(5)2 = (13)2 – b2
⇒ 25 = 169 – b2
⇒ b2 = 169 – 25 = 144
Put the values of a2 = 169 and b2 = 144 in equation
(i), we get, the equation of ellipse is
$$\frac{x^2}{169}+\frac{y^2}{144}=1.$$
16. Length of minor axis 16, foci (0, ± 6).
Sol. Since foci (0, ± 6) lies on Y-axis, as x-coordinate is zero. Hence, equation of ellipse will be of the form
$$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\space\text{...(i)}$$
Given that length of minor axis 2b = 16 and c = 5
⇒ b = 8 and c = 6
∵ c2 = a2 – b2
⇒ (6)2 = a2 – (8)2
⇒ 36 = a2 – 64
⇒ a2 = 36 + 64 = 100
Put the value of a2 = 100 and b2 = 64 in equation (i), we get, the equation of ellipse is
$$\frac{x^2}{64}+\frac{y^2}{100}=1.$$
17. Foci (± 3, 0), a = 4.
Sol. Since, foci (± 3, 0) lies on X-axis, as y-coordinate is zero.
Hence, equation of ellipse will be for the form
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\space\text{...(i)}$$
Given that foci (± c, 0) = (± 3, 0)
⇒ c = 3 and a = 4
∵ c2 = a2 – b2
⇒ (3)2 = (4)2 – b2
⇒ 9 = 16 – b2
⇒ b2 = 16 –9
⇒ b2 = 7
Put the value of a2 = 16 and b2 = 7 in equation (i), we get, equation of ellipse is
$$\frac{x^2}{16}+\frac{y^2}{7}=1.$$
18. b = 3, c = 4, centre of origin, foci on the X-axis.
Sol. Since, foci lies on X-axis, centre is at the origin. Hence, equation of ellipse will be of the form
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\space\text{...(i)}$$
Given that b = 3, c = 3
∵ c2 = a2 – b2
⇒ (4)2 = a2 – (3)2
⇒ 16 = a2 – 9
⇒ a2 = 16 + 9
⇒ a2 = 25
Put the values of a2 = 25 and b2 = 9 in equation (i),
we get, the equation of ellipse is
$$\frac{x^2}{25}+\frac{y^2}{9}=1.$$
19. Centre at (0, 0) major axis on the Y-axis passes through the points (3, 2) and (1, 6).
Sol. Since, major axis is along Y-axis and centre is at (0, 0).
Hence, equation of ellipse will be of the form
$$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\space\text{...(i)}$$
Given that equation (i) passes through the points (3, 2) and (1, 6) i.e., they will satisfy it.
$$\frac{3^2}{b^2}+\frac{2^2}{a^2}=1\\\Rarr\space\frac{9}{b^2}+\frac{4}{a^2}=1\space\text{...(ii)}\\\text{and}\space\frac{1^2}{b^2}+\frac{6^2}{a^2}=1\\\Rarr\space\frac{1}{b^2}+\frac{36}{a^2}=1\space\text{...(iii)}$$
Multiplying equation (ii) by 9 and then subtract the equation (iii) from it, we get
$$\frac{80}{b^2}=8\\\Rarr\space b^2=\frac{80}{8}$$
⇒ b2 = 10
From equation (ii),
$$\frac{9}{10}+\frac{4}{a^2}=1\\\Rarr\space \frac{4}{a^2}=1-\frac{9}{10}\\\Rarr\space\frac{4}{a^2}=\frac{1}{10}$$
⇒ a2 = 40
Put the values of a2 = 40 and b2 = 10 in equation (i), we get, the equation of ellipse is
$$\frac{x^2}{10}+\frac{y^2}{40}=1.$$
20. Major axis on the X-axis and passes through the points (4, 3) and (6, 2).
Sol. Since, major axis is along X-axis. Hence, equation of ellipse will be fo the form
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\space\text{...(i)}$$
Given that equation (i) passes through the points (4, 3) and (5, 2) i.e., they will satisfy it.
$$\therefore\space\frac{(4)^2}{a^2}+\frac{(3)^2}{b^2}=1\\\Rarr\space\frac{16}{a^2}+\frac{9}{b^2}=1\space\text{...(ii)}\\\text{and}\space\frac{(6)^2}{a^2}+\frac{(2)^2}{b^2}=1\\\Rarr\space\frac{36}{a^2}+\frac{4}{b^2}=1\space\text{...(iii)}$$
Multiplying equation (ii) by 4 and equation (iii) by 9, then subtracting, we get
$$\frac{64}{a^2}-\frac{324}{a^2}=4-9\\\Rarr\space -\frac{260}{a^2}=-5\\\Rarr\space a^2=\frac{260}{5}$$
⇒ a2 = 52
From equation (ii),
$$\frac{9}{b^2}=1-\frac{16}{52}\\\Rarr\space\frac{9}{b^2}=\frac{52-16}{52}\\\Rarr\space\frac{9}{b^2}=\frac{36}{52}\\\Rarr\space b^2=\frac{9×52}{36}=13$$
Put the values of a2 = 52 and b2 = 13 i equation (i), we get, the equation of ellipse is
$$\frac{x^2}{52}+\frac{y^2}{13}=1.$$
Exercise 11.4
Question (Q. Nos. 1 to 6): In each of the question, find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.
$$\textbf{1.}\space\frac{\textbf{x}^\textbf{2}}{\textbf{16}}-\frac{\textbf{y}^\textbf{2}}{\textbf{9}}\textbf{=1.}$$
Sol. Comparing the given equation
$$\frac{x^2}{16}-\frac{y^2}{9}=1\\\text{with}\space\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\text{we get}$$
⇒ a = 4 and b = 3
Now, c2 = a2 + b2 = 16 + 9 = 25
⇒ c = 5 (∵ c must be positive)
∴ The coordinate of foci = (± c, 0) = (± 5, 0)
The coordinate of vertices = (± a, 0), = (± 4, 0)
$$\text{Eccentricity e =}\frac{c}{a}=\frac{5}{4}\\\text{The length of latus rectum =}\frac{2b^2}{a}\\=\frac{2×9}{4}=\frac{9}{2}.$$
$$\textbf{2.}\space\frac{\textbf{y}^\textbf{2}}{\textbf{9}}-\frac{\textbf{x}^\textbf{2}}{\textbf{27}}\textbf{=1.}$$
Sol. Comparing the given equation
$$\frac{y^2}{9}-\frac{x^2}{27}=1\space\\\text{with}\space\frac{y^2}{a^2}-\frac{x^2}{b^2}=1,$$
we get a2 = 9 and b2 = 27
$$\Rarr\space\text{a = 3 and b=3}\sqrt{3}$$
(∵ a, b > 0)
∵ c2 = a2 + b2 = 9 + 27 = 36
⇒ c = 6 (∵ c must be positive)
∴ The coordinate of Foci = (0, ± c) = (0, ± 6)
The coordinate of vertices = (0, ± a) = (0, ± 3)
$$\text{Eccentricity e =}\frac{c}{a}=\frac{6}{3}=2\\\text{The length of latus rectum =}\frac{2b^2}{a}\\=\frac{2×27}{3}=18.$$
3. 9y2 – 4x2 = 36.
Sol. Given equation is 9y2 – 4x2 = 36, divide it by 36, we get
$$\frac{9y^2}{36}-\frac{4x^2}{36}=\frac{36}{36}\\\Rarr\space \frac{y^2}{4}-\frac{x^2}{9}=1\\\text{Now, comparing}\space\frac{y^2}{4}-\frac{x^2}{9}=1\\\text{with}\space\frac{y^2}{a^2}-\frac{x^2}{b^2}=1,$$
we get
a2 = 4 and b2 = 9
⇒ a = 2 and b = 3 (∵ a, b > 0)
∵ c2 = a2 + b2
⇒ c2 = 4 + 9 = 13
$$\Rarr\space c=\sqrt{13}\space(\because c\gt 0)$$
$$\therefore\space\text{The coordinate of foci = (0, ± c) =}\\(0,\pm\sqrt{13})$$
The coordinate of vertices = (0, ± a) = (0, ± 2)
$$\text{Eccentricity, e =}\frac{c}{a}=\frac{\sqrt{13}}{2}\\\text{The length of latus rectum =}\frac{2b^2}{a}\\=\frac{2×9}{2}=9.$$
4. 16x2 – 9y2 = 576.
Sol. Given equation is 16x2 – 9y2 = 576, divide it by 576, we get
$$\frac{16x^2}{576}-\frac{9y^2}{576}=\frac{576}{576}\\\Rarr\space\frac{x^2}{36}-\frac{y^2}{64}=1\\\text{Now, comparing}\space\frac{x^2}{36}-\frac{y^2}{64}=1\\\text{with}\space \frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$$
we get
a2 = 36 and b2 = 64
⇒ a = 6 and b = 8
∵ c2 = a2 + b2 = 36 + 64 = 100
⇒ c = 10
∴ The coordinate of foci = (± c, 0) = (± 10, 0)
The coordinate of vertices = (± a, 0) = (± 6, 0)
$$\text{Eccentricity, e =}\frac{c}{a}\\=\frac{10}{6}=\frac{5}{3}\\\text{The length of latus rectum =}\frac{2b^2}{a}\\=\frac{2×64}{6}=\frac{64}{3}.$$
5. 5y2 – 9x2 = 36.
Sol. Given equation is 5y2 – 9x2 = 36, divide it by 36, we get
$$\frac{5y^2}{36}-\frac{9x^2}{36}=\frac{36}{36}\\\Rarr\space\frac{y^2}{\frac{36}{5}}-\frac{x^2}{4}=1\\\text{On comparing}\space\frac{y^2}{\frac{36}{5}}-\frac{x^2}{4}=1\\\text{with}\space\frac{y^2}{a^2}-\frac{x^2}{b^2}=1,\\\text{we get}\\ a^2=\frac{36}{5}\text{and}\space b^2=4\\\Rarr\space a=\frac{6}{\sqrt{5}}\text{and b=2}\\\because\space c^2=a^2+b^2=\frac{36}{5}+4$$
$$\Rarr\space c^2=\frac{36+20}{5}=\frac{56}{5}\\\Rarr\space c=\frac{2\sqrt{14}}{\sqrt{5}}\\\therefore\space\text{The coordinate of Foci = (0, ± c) =}\\\bigg(0,\pm\frac{2\sqrt{14}}{\sqrt{5}}\bigg)\\\text{The coordinate of Vertices = (0, ± a) =}\\\bigg(0,\pm\frac{6}{\sqrt{5}}\bigg)\\\text{Eccentricity, e =}\frac{c}{a}\\=\frac{\frac{2\sqrt{14}}{\sqrt{5}}}{\frac{6}{\sqrt{5}}}=\frac{\sqrt{14}}{3}$$
$$\text{The length of Latus rectum =}\frac{2b^2}{a}\\=\frac{2×4}{\frac{6}{\sqrt{5}}}=\frac{8\sqrt{5}}{6}=\frac{4\sqrt{5}}{3}.$$
6. 49y2 – 16x2 = 784.
Sol. Given equation is 49y2 – 16x2 = 784, divide it by 784, we get
$$\frac{49y^2}{784}-\frac{16x^2}{784}=\frac{784}{784}\\\Rarr\space \frac{y^2}{16}-\frac{x^2}{49}=1\\\text{On comparing}\space\frac{y^2}{16}-\frac{x^2}{49}=1\\\text{with}\space\frac{y^2}{a^2}-\frac{x^2}{b^2}=1,\text{we get}$$
a2 = 16 and b2 = 49
⇒ a = 4 and b = 7
∵ c2 = a2 + b2 = 16 + 49 = 65
$$\Rarr\space c=\sqrt{65}\space (\because\space c\gt 0)\\\therefore\space\text{The corodinate of Foci}\\ = (0, ± c) =(0,\pm\sqrt{65})$$
The coordinate of Vertices = (0, ± a) = (0, ± 4)
$$\text{Eccentricity, e =}\frac{c}{a}=\frac{\sqrt{65}}{4}\\\text{The length of Latus rectum =}\frac{2b^2}{a}\\=\frac{2×49}{4}=\frac{49}{2}.$$
Direction (Q. Nos. 7 to 15): In each of the question find the equations of the hyperbola satisfying the given conditions.
7. Vertices (± 2, 0) and foci (± 3, 0).
Sol. Since, vertices (± 2, 0) and foci (± 3, 0) lies on X-axis, as coefficient of Y-axis is zero.
Hence, equation of hperbola will be of the form
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\space\text{...(i)}$$
where it is given that
∴ Foci = (± 3, 0)
Vertices = (± 2, 0)
⇒ a = 2 and c = 3
∵ c2 = a2 + b2
⇒ 9 = 4 + b2
⇒ b2 = 5
Put the values of a2 = 4 and b2 = 5 in equation (i), we get, the equation of hyperbola is
$$\frac{x^2}{4}-\frac{y^2}{5}=1.$$
8. Vertices (0, ± 5), foci (0, ± 8).
Sol. Since, vertices (0, ± 5) and foci (0, ± 8) lies on Y-axis, as coefficient of X-axis is zero.
Hence, equation of hyperbola will be of the form
$$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\space\text{...(i)}$$
where it is given that vertices (0, ± a) = (0, ± 5) and foci (0, ± c) = (0, ± 8)
a = 5 and c = 8
∵ c2 = a2 + b2
⇒ 64 = 25 + b2
⇒ b2 = 64 – 25
⇒ b2 = 39
Put a2 = 25 and b2 = 39 in equation (i), we get
∴ The equation of hyperbola is
$$\frac{y^2}{25}-\frac{x^2}{39}=1.$$
9. Vertices (0, ± 3) and foci (0, ± 5).
Sol. Since, vertices (0, ± 3) and foci (0, ± 5) lies on Y-axis, as coordiante of X is zero.
Hence, equation of hperbola will be of the form
$$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\space\text{...(i)}$$
where it is given that vertices (0, ± 3) = (0, ± a) and foci (0, ± 5) = (0, ± c)
⇒ a = 3 and c = 5
∵ c2 = a2 + b2
⇒ 25 = 9 + b2
⇒ b2 = 16
Put a2 = 9 and b2 = 16 in equation (i), we get
The equation of hyperbola
$$\frac{y^2}{9}-\frac{x^2}{16}=1.$$
10. Foci (± 5, 0) and the tranverse axis is length 8.
Sol. Since, foci (± 5, 0) lies on X-axis, as coordinate of Y is zero.
Hence, equation of hyperbola will be of the form
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\space\text{...(i)}$$
where it is given that foci (± 5, 0) = (± c, 0) and
length of transverse axis if 2a = 8.
c = 5 and 2a = 8
⇒ a = 4
∵ c2 = a2 + b2
⇒ 25 = 16 + b2
⇒ b2 = 9
Put a2 = 16 and b2 = 9 in equation (i), we get
The equation of hyperbola is
$$\frac{x^2}{16}-\frac{y^2}{9}=1.$$
11. Foci (0, ± 13), the conjugate axis is of length 24.
Sol. Since, foci (0, ± 13) lies on Y-axis as x-coordinate is zero. Hence, equation of hperbola will be of the form.
$$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\space\text{...(i)}$$
where it is given that foci (0, ± 13) = (0, ± c) and conjugate length, 2b = 24
⇒ c = 13 and b = 12
∵ c2 = a2 + b2
∴ 169 = a2 + 144
⇒ a2 = 169 – 144
⇒ a2 = 25
Put a2 = 25 and b2 = 144 in equation (i), we get
∴ The equation of hyperbola is
$$\frac{y^2}{25}-\frac{x^2}{144}=1.$$
$$\textbf{12. Foci }(\pm \textbf{3}\sqrt{\textbf{5}}\textbf{, 0})\\\textbf{the latus rectum is of length 8.}\\\textbf{Sol.\space}\text{Since, foci}\space(\pm 3\sqrt{5},0)\space\text{lies on X-axis,}$$
as y-coordinate is zero. Thus, equation of hperbola will be of the form
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\space\text{...(i)}$$
where it is given that foci
$$(\pm 3\sqrt{5},0)=(\pm c,0)\\ c= 3\sqrt{5}$$
and length of latusrectum
$$\frac{2b^2}{a}=8$$
⇒ b2 = 4a
∵ c2 = a2 + b2
$$\therefore\space(3\sqrt{5})^2=a^2+(4a)$$
⇒ 9 × 5 = a2 + 4a
⇒ a2 + 4a – 45 = 0
On splitting the middle term
⇒ a2 + 9a – 5a – 45 = 0
On splitting the middle term
⇒ a2 + 9a – 5a – 45 = 0
⇒ a(a + 9) – 5(a + 9) = 0
⇒ (a – 5) (a + 9) = 0
⇒ a – 5 = 0, a + 9 = 0
⇒ a = 5, – 9
∵ a can’t be negative.
So, a = 5
⇒ b2 = 4 × 5 = 20
Put a2 = 25 and b2 = 20 in equation (i), we get
∴ The equation of hyperbola is
$$\frac{x^2}{25}-\frac{y^2}{20}=1.$$
13. Foci (± 4, 0) the latus rectum is of length 12.
Sol. Since foci (± 4, 0) lies on X-axis. Hence, euation of hyperbola will be of the form
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\space\text{...(i)}$$
where it is given that foci (± 4, 0) = (± c, 0) and length of latusrectum,
$$\frac{2b^2}{a}=12$$
∴ c = 4 and b2 = 6a …(ii)
∵ c2 = a2 + b2
⇒ 16 = a2 + 6a
⇒ a2 + 6a – 16 = 0
Splitting the middle term
a2 + 8a – 2a – 16 = 0
⇒ a(a + 8) – 2(a + 8) = 0
⇒ (a – 2) (a + 8) = 0
⇒ a – 2 = 0, a + 8 = 0
⇒ a = 2, – 8
·.· a can’t be negative.
So, a = 2
⇒ b2 = 6a = 6 × 2 = 12
Put a2 = 4 and b2 = 12 in equation (i), we get
∴ The equation of hyperbola is
$$\frac{x^2}{4}-\frac{y^2}{12}=1.$$
$$\textbf{14. Vertices (± 7, 0) and e =}\frac{\textbf{4}}{\textbf{3}}.$$
Sol. Since, vertices (± 7, 0) lies on X-axis, as y-coordinate is zero. Hence, equation of hperbola will be of the form
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\space\text{...(i)}$$
where it is given that vertices (± 7, 0) = (± a, 0)
$$\Rarr\space\text{a=7}\space\text{and}\space e=\frac{4}{3}$$
∵ c = ae
$$\Rarr\space c=7×\frac{4}{3}=\frac{28}{3}$$
Also, c2 = a2 + b2
$$\frac{784}{9}=49+b^2\\\Rarr\space b^2=\frac{784}{9}-\frac{49}{1}\\\Rarr\space b^2=\frac{784-441}{9}\\\Rarr\space b^2=\frac{343}{9}$$
$$\text{Put}\space a^2=49\space\text{and}\space b^2=\frac{343}{9}\\\text{in equation (i), we get}\\\frac{x^2}{49}-\frac{y^2}{\frac{343}{9}}=1\\\therefore\space\text{The equation of hyperbola is}\\\frac{x^2}{49}-\frac{9y^2}{343}=1.$$
$$\textbf{15.}\space\textbf{Foci}\space(\textbf{0,}\pm\sqrt{\textbf{10}})\space \textbf{passing}\\\textbf{ through the point (2, 3).}$$
Sol. Since, foci lies on Y-axis, as x-coordinate in zero. Hence, equation of hyperbola will be of the form
$$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\space\text{...(i)}$$
where it is given that foci
$$(0,\pm\sqrt{10})=(0\pm c)\\\Rarr\space c=\sqrt{10}$$
Since the point (2, 3) lies on the hyperbola i.e., point will satisfy equation (i),
$$\therefore\space\frac{(3)^2}{a^2}-\frac{(2)^2}{b^2}=1\\\Rarr\space \frac{9}{a^2}-\frac{4}{b^2}=1\space\text{...(ii)}\\\because\space c=\sqrt{10}$$
⇒ c2 = 10
⇒ a2 + b2 = 10 (∵ c2 = a2 + b2)
⇒ b2 = 10 – a2 …(iii)
From equation (ii),
$$\frac{9}{a^2}-\frac{4}{10-a^2}=1\\\Rarr\space\frac{90-9a^2-4a^2}{a^2(10-a^2)}=1$$
⇒ 90 – 13a2 = 10a2 – a4
⇒ a4 – 23a2 + 90 = 0
⇒ a4 – 18a2 – 5a2 + 90 = 0
⇒ a2(a2 – 18) – 5(a2 – 18) = 0
⇒ (a2 – 5) (a2 – 18) = 0
⇒ a2 = 5 or a2 = 18
From equation (iii), if
a2 = 5
⇒ b2 = 10 – 5 = 5
If a2 = 18
⇒ b2 = 10 – 18 = – 8
Which is not possible.
So, a2 = b2 = 5
Put a2 = b2 = 5 in equation (i), we get
$$\Rarr\space\frac{y^2}{5}-\frac{x^2}{5}=1$$
⇒ y2 – x2 = 5
∴ The equation of hyperbla is
y2 – x2 = 5.
Miscellaneous Exercise
1. If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.
Sol. Taking vertex of the parabolic reflector at origin, X-axis along the axis of parabola. The equation of the parabola is y2 = 4ax. Given depth is 5 cm, diameter is 20 cm.
∵ Point P(5, 10) lies on parabola.
∴ (10)2 = 4a(5)
$$\Rarr\space a=\frac{100}{20}=5$$
∴ Focus (a, 0) = (5, 0)
Hence focus is the mid-point of given diameter. i.e.,
2. An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola.
Sol. Since, axis is vertical, so let arch of parabola is the form
x2 = 4ay …(i)
Given, OB = 10 m
and AC = 5 m
⇒ AB = (∵ AC = 5)
$$\text{Hence, coordinate at A =}\bigg(\frac{5}{2},10\bigg)\\\text{will satisfy equation (i).}\\\text{i.e.,}\space\bigg(\frac{5}{2}\bigg)^2=4a×10\\\Rarr\space\frac{25}{4}=40a\\\Rarr\space a=\frac{5}{32}$$
From equation (i),
$$x^2=4×\frac{5}{32}y$$
Now, let OR = 2 and PQ = k
$$\Rarr\space\text{RP}=\frac{k}{2}\\\text{Therefore, P =}\bigg(\frac{k}{2},2\bigg)\space\\\text{will lies on parabola.}\\\therefore\space\bigg(\frac{k}{2}\bigg)^2=\frac{5}{8}×2\\\bigg(\because\space x^2=\frac{5}{8}y\bigg)\\\Rarr\space\frac{k^2}{4}=\frac{5}{4}\\\Rarr\space k=\sqrt{5}=2.23\space\text{m (approx.)}$$
3. The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle.
Sol. Since, wires are vertical. Let equation of the parabola is in the form
x2 = 4ay …(i)
Focus is at the middle of the cable and shortest and longest vertical supports are 6 m and 30 m and roadway in 100 m long.
Clearly, coordinate of Q(50, 24) will satisfy equation (i)
(50)2 = 4a × 24
⇒ 2500 = 96a
$$\Rarr\space a=\frac{2500}{96}$$
Hence, from equation (i),
$$x^2=4×\frac{2500}{96}y\\\Rarr\space x^2=\frac{2500}{24}y$$
Let PR = k m
∴ Point P(18, k) will satisfy the equation of parabola i.e.,
From equation (i),
$$(18)^2=\frac{2500}{24}×k\\\Rarr\space 324=\frac{2500}{24}×k\\\Rarr\space k=\frac{324×24}{2500}\\=\frac{324×6}{625}=\frac{1944}{625}$$
k = 3.11
∴ Required length = 6 + k = 6 + 3.11 = 9.11 m (approx.)
4. An arch is in the form of a semi ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.
Sol.
Clearly, equation of ellipse takes the form
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\space\text{...(i)}$$
Here, it is given
2a = 8 and b = 2
⇒ a = 4, b = 2
Put the value of a and b in equation (i), we get
$$\frac{x^2}{16}+\frac{y^2}{4}=1$$
Given, AP = 1.5 m
⇒ OP = OA – AP = 4 – 1.5
⇒ OP = 2.5 m
Let PQ = k
∴ Coordinate Q = (2.5, k) will satisfy the equation of ellipse.
$$\text{i.e.,}\space\frac{(2.5)^2}{16}+\frac{k^2}{4}=1\\\Rarr\space \frac{6.25}{16}+\frac{k^2}{4}=1\\\Rarr\space \frac{k^2}{4}=\frac{1}{1}-\frac{6.25}{16}\\=\frac{16-6.25}{16}\\\Rarr\space\frac{k^2}{4}=\frac{9.75}{16}\\\Rarr\space k^2=\frac{9.75}{4}$$
⇒ k2 = 2.4375
⇒ k = 1.56 m (approx.)
Hence, the required height of arch = 1·56 m.
5. A rod of length 12 cm moves with it ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the X-axis.
Sol. Let AB be the road of length 12 cm.
Let OA = a and OB = b
Using by Pythagoras theorem ΔOAB.
(Base)2 + (Perpendicular)2
= (Hypotenuse)2
⇒ a2 + b2 = (12)2
⇒ a2 + b2 = 144 …(i)
Let AP = 3 cm
⇒ PB = AB – AB = 12 – 3 = 9 cm
⇒ AP : PB = 3 : 9 = 1 : 3
Let P = (h, k) for which locus to be found.
∴ If P(h, k) divides the points A and B internally in the ratio m : n, then
$$h=\frac{mx_2+nx_1}{m+n},\\k=\frac{my_2+ny_1}{m+n}\\\therefore\space h=\frac{1×0+3×a}{1+3}$$
$$a=\frac{4h}{3}\\\text{and}\space k=\frac{1×b+3×0}{1+3}\\k=\frac{b}{4}$$
⇒ b = 4k
Put the value of a =
$$\text{Put the value of a =}\frac{4k}{3}\\\text{and b=4k in equation (i).}\\\text{we get}\\\bigg(\frac{4h}{3}\bigg)^2+(4k)^2=144\\\Rarr\space\frac{16h^2}{9}+16k^2=144\\\Rarr\space\frac{h^2}{9}+k^2=9\\\Rarr\space \frac{h^2}{81}+\frac{k^2}{9}=1$$
Hence, locus of a point P(h, k) is
$$\frac{x^2}{81}+\frac{y^2}{9}=1.$$
Hence, the equation of the point P on the rod is
$$\frac{x^2}{81}+\frac{y^2}{9}=1.$$
6. Find the area of the triangle formed by the lines joining the vertex of the parabola x2 = 12y to the ends of its latusrectum.
Sol. Clearly, latus rectum is a line perpendicular to the axes and passing through focus whose length is 4a.
Given, x2 = 12y,
which of the form x2 = 4ay
⇒ 4a = 12 = Length of ABC
∴ Focus C = (0, 3)
$$\text{Area of} Δ\text{OAB =}\frac{1}{2}×AB×OC\\\bigg[\because \text{Area of D}=\frac{1}{2}×\text{base×height}\bigg]\\=\frac{1}{2}×12×3$$
= 6 × 3
= 18 sq. unit
7. A man running a race course notes that sum of its distances from two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.
Sol. Clearly path traced by the man will be ellipse
Given, SP + S′P = 10
i.e., 2a = 10
⇒ a = 5
Since, the coordinates of S and S′ are (c, 0) and (– c, 0). Therefore, distance between S and S′ is
2c = 8
⇒ c = 4
∵ c2 = a2 – b2
⇒ 16 = 25 – b2
⇒ b2 = 25 – 16
⇒ b2 = 9
⇒ b = 3
Hence, equation of path (ellipse) is
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\\Rarr\space\frac{x^2}{25}+\frac{y^2}{9}=1\\(\because a=5, b=3)$$
8. An equilateral triangle is inscribed in the parabola y2 = 4ax, where one vertex is at the vertex of the parabola. Find the lenght of the side of the triangle.
Sol. First, we draw the parabola in the positive side of X-axis and inside that draw an equilateral triangle OAB.
Let OB = l = OA = AB
∴ ∠BOA = 60°
⇒ ∠BOP = 30°
In ΔBOP,
$$\text{sin 30\degree}=\frac{\text{PB}}{\text{OB}}\\\Rarr\space\frac{1}{2}=\frac{\text{PB}}{l}\\\Rarr\space\text{PB}=\frac{l}{2}$$
$$\text{and cos 30\degree}=\frac{\text{OP}}{\text{OB}}\\\Rarr\space\frac{\sqrt{3}}{2}=\frac{\text{OP}}{\text{OB}}\\\Rarr\space\frac{\sqrt{3}}{2}=\frac{\text{OP}}{l}\\\Rarr\space\text{OP}=\frac{l\sqrt{3}}{2}\\\therefore\space\text{Coordinate of B = (OP, PB) =}\space\bigg(\frac{l\sqrt{3}}{2},\frac{l}{2}\bigg)\\\text{will satisfy y}^2 \text{= 4ax i.e.,}$$
$$\bigg(\frac{l}{2}\bigg)^2=\frac{4a×l\sqrt{3}}{2}\\\Rarr\space\frac{l^2}{4}=\frac{4al\sqrt{3}}{2}\\\Rarr\space l=8\sqrt{3a}\\\text{Hence, the length of side of the equilateral}\\\text{triangle is}\space8\sqrt{3}a.$$
NCERT Solutions for Class 11 Maths Chapter 10 Free PDF Download
Please Click on Free PDF Download link to Download the NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections
