# NCERT Solutions | Class 11 Maths Chapter 11 | Conic sections

## CBSE Solutions | Maths Class 11

Check the below NCERT Solutions for Class 11 Maths Chapter 11 Conic sections Pdf free download. NCERT Solutions Class 11 Maths  were prepared based on the latest exam pattern. We have Provided Conic sections Class 11 Maths NCERT Solutions to help students understand the concept very well.

### NCERT | Class 11 Maths

Book: National Council of Educational Research and Training (NCERT) Central Board of Secondary Education (CBSE) 11th Maths 11 Conic sections English

#### Conic sections | Class 11 Maths | NCERT Books Solutions

You can refer to MCQ Questions for Class 11 Maths Chapter 11 Conic sections to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

## NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1

In each of the following Exercises 1 to 5, find the equation of the circle with

Ex 11.1 Class 11 Maths Question 1.

centre (0, 2) and radius 2
Solution.
Here h = 0,k = 2 and r = 2
The equation of circle is,
(x-h)2 + (y- k)2 = r2
∴ (x – 0)2 + (y – 2)2 = (2)2
⇒ x2 + y2 + 4 – 4y = 4
⇒ x2 + y2 – 4y = 0

Ex 11.1 Class 11 Maths Question 2.

Solution.
Here h=-2,k = 3 and r = 4
The equation of circle is,
(x – h)2 + (y – k)2 = r2
∴(x + 2)2 + (y – 3)2 = (4)2
⇒ x2 + 4 + 4x + y2 + 9 – 6y = 16
⇒ x2 + y2 + 4x – 6y – 3 = 0

Ex 11.1 Class 11 Maths Question 3.

centre $$\left( \frac { 1 }{ 2 } ,\quad \frac { 1 }{ 4 } \right)$$ and radius $$\frac { 1 }{ 12 }$$
Solution.
here h = $$\frac { 1 }{ 2 }$$, k = $$\frac { 1 }{ 4 }$$ and r = $$\frac { 1 }{ 12 }$$
The equation of circle is,

Ex 11.1 Class 11 Maths Question 4.

centre (1, 1) and radius $$\sqrt { 2 }$$
Solution.
Here h = l, k=l and r = $$\sqrt { 2 }$$
The equation of circle is,
(x – h)2 + (y – k)2 = r2
∴ (x – 1)2 + (y – 1)2 = $$\left( \sqrt { 2 } \right)$$2
⇒ x2 + 1 – 2x + y2 +1 – 2y = 2
⇒ x2 + y2 – 2x – 2y = 0

Ex 11.1 Class 11 Maths Question 5.

centre (-a, -b) and radius $$\sqrt { { a }^{ 2 }-{ b }^{ 2 } }$$.
Solution.
Here h=-a, k = -b and r = $$\sqrt { { a }^{ 2 }-{ b }^{ 2 } }$$
The equation of circle is, (x – h)2 + (y – k)2 = r2
∴ (x + a)2 + (y + b)2 = $$\left( \sqrt { { a }^{ 2 }-{ b }^{ 2 } } \right)$$
⇒ x2 + a2 + 2ax + y2 + b2 + 2by = a2 -b2
⇒ x2 + y2 + 2ax + 2 by + 2b2 = 0

In each of the following exercises 6 to 9, find the centre and radius of the circles.

Ex 11.1 Class 11 Maths Question 6.

(x + 5)2 + (y – 3)2 = 36
Solution.
The given equation of circle is,
(x + 5)2 + (y – 3)2 = 36
⇒ (x + 5)2 + (y – 3)2 = (6)2
Comparing it with (x – h)22 + (y – k)2 = r2, we get
h = -5, k = 3 and r = 6.
Thus the co-ordinates of the centre are (-5, 3) and radius is 6.

Ex 11.1 Class 11 Maths Question 7.

x2 + y2 – 4x – 8y – 45 = 0
Solution.
The given equation of circle is
x2 + y2 – 4x – 8y – 45 = 0
∴ (x2 – 4x) + (y2 – 8y) = 45
⇒ [x2 – 4x + (2)2] + [y2 – 8y + (4)2] = 45 + (2)2 + (4)2
⇒ (x – 2)2 + (y – 4)2 = 45 + 4 + 16
⇒ (x – 2)2 + (y – 4)2 = 65
⇒ (x – 2)2 + (y – 4)2= $$\left( \sqrt { 65 } \right) ^{ 2 }$$
Comparing it with (x – h)2 + (y – k)2 = r2, we
have h = 2,k = 4 and r = $$\sqrt { 65 }$$.
Thus co-ordinates of the centre are (2, 4) and radius is $$\sqrt { 65 }$$.

Ex 11.1 Class 11 Maths Question 8.

x2 + y2 – 8x + 10y – 12 = 0
Solution.
The given equation of circle is,
x2 + y2 – 8x + 10y -12 = 0
∴ (x2 – 8x) + (y2 + 10y) = 12
⇒ [x2 – 8x + (4)2] + [y2 + 10y + (5)2] = 12 + (4)2 + (5)2
⇒ (x – 4)2 + (y + 5)2 = 12 + 16 + 25
⇒ (x – 4)2 + (y + 5)2 = 53
⇒ (x – 4)2 + (y + 5)2 = $$\left( \sqrt { 53 } \right) ^{ 2 }$$
Comparing it with (x – h)2 + (y – k)2 = r2, we have h = 4, k = -5 and r = $$\sqrt { 53 }$$
Thus co-ordinates of the centre are (4, -5) and radius is $$\sqrt { 53 }$$.

Ex 11.1 Class 11 Maths Question 9.

2x2 + 2y2 – x = 0
Solution.
The given equation of circle is,
2x2 + 2y2 – x = 0

Ex 11.1 Class 11 Maths Question 10.

Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.
Solution.
The equation of the circle is,
(x – h)2 + (y – k)2 = r2 ….(i)
Since the circle passes through point (4, 1)
∴ (4 – h)2 + (1 – k)2 = r2
⇒ 16 + h2 – 8h + 1 + k2 – 2k = r2
⇒ h2+ k2 – 8h – 2k + 17 = r2 …. (ii)
Also, the circle passes through point (6, 5)
∴ (6 – h2 + (5 – k)2 = r2
⇒ 36 + h2 -12h + 25 + k2 – 10k = r2
⇒ h2 + k2 – 12h – 10kk + 61 = r2 …. (iii)
From (ii) and (iii), we have h2 + k2 – 8h – 2k +17
= h2 + k2– 12h – 10k + 61
⇒ 4h + 8k = 44 => h + 2k = ll ….(iv)
Since the centre (h, k) of the circle lies on the line 4x + y = 16
∴ 4h + k = 16 …(v)
Solving (iv) and (v), we get h = 3 and k = 4.
Putting value of h and k in (ii), we get
(3)2 + (4)2 – 8 x 3 – 2 x 4 + 17 = r2
∴ r2 = 10
Thus required equation of circle is
(x – 3)2 + (y – 4)2 = 10
⇒ x2 + 9 – 6x + y2 +16 – 8y = 10
⇒ x2 + y2 – 6x – 8y +15 = 0.

Ex 11.1 Class 11 Maths Question 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.
Solution.
The equation of the circle is,
(x – h)2 + (y – k)2 = r2 ….(i)
Since the circle passes through point (2, 3)
∴ (2 – h)2 + (3 – k)2 = r2
⇒ 4 + h2 – 4h + 9 + k2 – 6k = r2
⇒ h2+ k2 – 4h – 6k + 13 = r2 ….(ii)
Also, the circle passes through point (-1, 1)
∴ (-1 – h)2 + (1 – k)2 = r2
⇒ 1 + h2 + 2h + 1 + k2 – 2k = r2
⇒ h2 + k2 + 2h – 2k + 2 = r2 ….(iii)
From (ii) and (iii), we have
h2 + k2 – 4h – 6k + 13 = h2 + k2 + 2h – 2k + 2
⇒ -6h – 4k = -11 ⇒ 6h + 4k = 11 …(iv)
Since the centre (h, k) of the circle lies on the line x – 3y-11 = 0.
∴ h – 3k – 11 = 0 ⇒ h -3k = 11 …(v)
Solving (iv) and (v), we get
h = $$\frac { 7 }{ 2 }$$ and k = $$\frac { -5 }{ 2 }$$
Putting these values of h and k in (ii), we get
$$\left( \frac { 7 }{ 2 } \right) ^{ 2 }+\left( \frac { -5 }{ 2 } \right) ^{ 2 }-\frac { 4\times 7 }{ 2 } -6\times \frac { -5 }{ 2 } +13={ r }^{ 2 }$$
⇒ $$\frac { 49 }{ 4 } +\frac { 25 }{ 4 } -14+15+13$$ ⇒ $${ r }^{ 2 }=\frac { 65 }{ 2 }$$
Thus required equation of circle is
⇒ $$\left( x-\frac { 7 }{ 2 } \right) ^{ 2 }+\left( y+\frac { 5 }{ 2 } \right) ^{ 2 }=\frac { 65 }{ 2 }$$
⇒ $${ x }^{ 2 }+\frac { 49 }{ 4 } -7x+{ y }^{ 2 }+\frac { 25 }{ 4 } +5y=\frac { 65 }{ 2 }$$
⇒ 4x2 + 49 – 28x + 4y2 + 25 + 20y = 130
⇒ 4x2 + 4y2 – 28x + 20y – 56 = 0
⇒ 4(x2 + y2 – 7x + 5y -14) = 0
⇒ x2 + y2 – 7x + 5y -14 = 0.

Ex 11.1 Class 11 Maths Question 12.

Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).
Solution.
Since the centre of the circle lies on x-axis, the co-ordinates of centre are (h, 0). Now the circle passes through the point (2, 3).

Ex 11.1 Class 11 Maths Question 13.

Find the equation of the circle passing through (0, 0) and making intercepts a and b on the co-ordinate axes.
Solution.
Let the circle makes intercepts a with x-axis and b with y-axis.
∴ OA = a and OB = b
So the co-ordinates of A are (a, 0) and B are (0,b)
Now, the circle passes through three points 0(0, 0), A(a, 0) and B(0, b).

Ex 11.1 Class 11 Maths Question 14.

Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).
Solution.
The equation of circle is
(x – h)2 + (y – k)2 = r2 ….(i)
Since the circle passes through point (4, 5) and co-ordinates of centre are (2, 2)

Ex 11.1 Class 11 Maths Question 15.

Does the point (-2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?
Solution.
The equation of given circle is x2 + y2 = 25
⇒ (x – 0)2 + (y – 0)2 = (5)2
Comparing it with (x – h)2 + (y – k)2 = r2, we
get
h = 0,k = 0, and r = 5
Now, distance of the point (-2.5, 3.5) from the centre (0, 0)
$$\sqrt { \left( 0+2.5 \right) ^{ 2 }+\left( 0-3.5 \right) ^{ 2 } } =\sqrt { 6.25+12.25 }$$
$$\sqrt { 18.5 }$$ = 4.3 < 5.
Thus the point (-2.5, 3.5) lies inside the circle.

## NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.2

In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

Ex 11.2 Class 11 Maths Question 1.

y2= 12x
Solution.
The given equation of parabola is y2 = 12x which is of the form y2 = 4ax.
∴ 4a = 12 ⇒ a = 3
∴ Coordinates of focus are (3, 0)
Axis of parabola is y = 0
Equation of the directrix is x = -3 ⇒ x + 3 = 0
Length of latus rectum = 4 x 3 = 12.

Ex 11.2 Class 11 Maths Question 2.

x2 = 6y
Solution.
The given equation of parabola is x2 = 6y which is of the form x2 = 4ay.

Ex 11.2 Class 11 Maths Question 3.

y2 = – 8x
Solution.
The given equation of parabola is
y2 = -8x, which is of the form y2 = – 4ax.
∴ 4a = 8 ⇒ a = 2
∴ Coordinates of focus are (-2, 0)
Axis of parabola is y = 0
Equation of the directrix is x = 2 ⇒ x – 2 = 0
Length of latus rectum = 4 x 2 = 8.

Ex 11.2 Class 11 Maths Question 4.

x2 = -16y
Solution.
The given equation of parabola is
x2 = -16y, which is of the form x2 = -4ay.
∴ 4a = 16 ⇒ a = 4
∴ Coordinates of focus are (0, -4)
Axis of parabola is x = 0
Equation of the directrix is y = 4 ⇒ y – 4 = 0
Length of latus rectum = 4 x 4 = 16.

Check the Parabola Calculator to solve Parabola Equation.

Ex 11.2 Class 11 Maths Question 5.

y2= 10x
Solution.
The given equation of parabola is y2 = 10x, which is of the form y2 = 4ax.

Ex 11.2 Class 11 Maths Question 6.

x2 = -9y
Solution.
The given equation of parabola is
x2 = -9y, which is of the form x2 = -4ay.

In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:

Ex 11.2 Class 11 Maths Question 7.

Focus (6, 0); directrix x = -6
Solution.
We are given that the focus (6, 0) lies on the x-axis, therefore x-axis is the axis of parabola. Also, the directrix is x = -6 i.e. x = -a and focus (6, 0) i.e. (a, 0). The equation of parabola is of the form y2 = 4ax.
The required equation of parabola is
y2 = 4 x 6x ⇒ y2 = 24x.

Ex 11.2 Class 11 Maths Question 8.

Focus (0, -3); directri xy=3
Solution.
We are given that the focus (0, -3) lies on the y-axis, therefore y-axis is the axis of parabola. Also the directrix is y = 3 i.e. y = a and focus (0, -3) i.e. (0, -a). The equation of parabola is of the form x2 = -4ay.
The required equation of parabola is
x2 = – 4 x 3y ⇒ x2 = -12y.

Ex 11.2 Class 11 Maths Question 9.

Vertex (0, 0); focus (3, 0)
Solution.
Since the vertex of the parabola is at (0, 0) and focus is at (3, 0)
∴ y = 0 ⇒ The axis of parabola is along x-axis
∴ The equation of the parabola is of the form y2 = 4ax
The required equation of the parabola is
y2 = 4 x 3x ⇒ y2 = 12x.

Ex 11.2 Class 11 Maths Question 10.

Vertex (0, 0); focus (-2, 0)
Solution.
Since the vertex of the parabola is at (0, 0) and focus is at (-2, 0).
∴ y = 0 ⇒ The axis of parabola is along x-axis
∴ The equation of the parabola is of the form y2 = – 4ax
The required equation of the parabola is
y2 = – 4 x 2x ⇒ y2 = -8x.

Ex 11.2 Class 11 Maths Question 11.

Vertex (0, 0), passing through (2, 3) and axis is along x-axis.
Solution.
Since the vertex of the parabola is at (0, 0) and the axis is along x-axis.
∴ The equation of the parabola is of the form y2 = 4ax
Since the parabola passes through point (2, 3)

Ex 11.2 Class 11 Maths Question 12.

Vertex (0, 0), passing through (5, 2) and symmetric with respect toy-axis.
Solution.
Since the vertex of the parabola is at (0, 0) and it is symmetrical about the y-axis.
∴ The equation of the parabola is of the form x2 = 4ay
Since the parabola passes through point (5, 2)

## NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3

In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

Ex 11.3 Class 11 Maths Question 1.

$$\frac { { x }^{ 2 } }{ 36 } +\frac { { y }^{ 2 } }{ 16 } =1$$
Solution.
Given equation of ellipse of $$\frac { { x }^{ 2 } }{ 36 } +\frac { { y }^{ 2 } }{ 16 } =1$$
Clearly, 36 > 16
The equation of ellipse in standard form is

Ex 11.3 Class 11 Maths Question 2.

$$\frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 25 } =1$$
Solution.
Given equation of ellipse is $$\frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 25 } =1$$
Clearly, 25 > 4
The equation of ellipse in standard form is

Ex 11.3 Class 11 Maths Question 3.

$$\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1$$
Solution.
Given equation of ellipse is $$\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1$$
Clearly, 16 > 9
The equation of ellipse in standard form is

Ex 11.3 Class 11 Maths Question 4.

$$\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 100 } =1$$
Solution.
Given equation of ellipse is $$\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 100 } =1$$
Clearly, 100 > 25
The equation of ellipse in standard form is

Ex 11.3 Class 11 Maths Question 5.

$$\frac { { x }^{ 2 } }{ 49 } +\frac { { y }^{ 2 } }{ 36 } =1$$
Solution.
Given equation of ellipse is $$\frac { { x }^{ 2 } }{ 49 } +\frac { { y }^{ 2 } }{ 36 } =1$$
Clearly, 49 > 36
The equation of ellipse in standard form is

Ex 11.3 Class 11 Maths Question 6.

$$\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1$$
Solution.
Given equation of ellipse is $$\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1$$
Clearly, 400 > 100
The equation of ellipse in standard form is

Ex 11.3 Class 11 Maths Question 7.

36x2 + 4y2 = 144
Solution.
Given equation of ellipse is 36x2 + 4y2 = 144

Ex 11.3 Class 11 Maths Question 8.

16x2 + y2 = 16
Solution.
Given equation of ellipse is16x2 + y2 = 16

Ex 11.3 Class 11 Maths Question 9.

4x2 + 9y2 = 36
Solution.
Given equation of ellipse is4x2 + 9y2 = 36

In each 0f the following Exercises 10 to 20, find the equation for the ellipse that satisfies the given conditions:

Ex 11.3 Class 11 Maths Question 10.

Vertices (±5, 0), foci (±4,0)
Solution.
Clearly, The foci (±4, o) lie on x-axis.
∴ The equation of ellipse is standard form is

Ex 11.3 Class 11 Maths Question 11.

Vertices (0, ±13), foci (0, ±5)
Solution.
Clearly, The foci (0, ±5) lie on y-axis.
∴ The equation of ellipse is standard form is

Ex 11.3 Class 11 Maths Question 12.

Vertices (±6, 0), foci (±4,0)
Solution.
Clearly, The foci (±4, 0) lie on x-axis.
∴ The equation of ellipse is standard form is

Ex 11.3 Class 11 Maths Question 13.

Ends of major axis (±3, 0), ends of minor axis (0, ±2)
Solution.
Since, ends of major axis (±3, 0) lie on x-axis.
∴ The equation of ellipse in standard form

Ex 11.3 Class 11 Maths Question 14.

Ends of major axis (0, $$\pm \sqrt { 5 }$$), ends of minor axis (±1, 0)
Solution.
Since, ends of major axis (0, $$\pm \sqrt { 5 }$$) lie on i-axis.
∴ The equation of ellipse in standard form

Ex 11.3 Class 11 Maths Question 15.

Length of major axis 26, foci (±5, 0)
Solution.
Since the foci (±5, 0) lie on x-axis.
∴ The equation of ellipse in standard form

Ex 11.3 Class 11 Maths Question 16.

Length of major axis 16, foci (0, ±6)
Solution.
Since the foci (0, ±6) lie on y-axis.
∴ The equation of ellipse in standard form

Ex 11.3 Class 11 Maths Question 17.

Foci (±3, 0) a = 4
Solution.
since the foci (±3, 0) on x-axis.
∴ The equation of ellipse in standard form

Ex 11.3 Class 11 Maths Question 18.

b = 3, c = 4, centre at the origin; foci on the x axis.
Solution.
Since the foci lie on x-axis.
∴ The equation of ellipse in standard form is

Ex 11.3 Class 11 Maths Question 19.

Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6)
Solution.
Since the major axis is along y-axis.
∴ The equation of ellipse in standard form

Ex 11.3 Class 11 Maths Question 20.

Major axis on the x-axis and passes through the points (4, 3) and (6, 2).
Solution.
Since the major axis is along the x-axis.
∴ The equation of ellipse in standard form is

## NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.4

In each of the Exercises 1 to 6, find the coordinates of the foci and the vertices, eccentricity and the length of the latus rectum of the hyperbolas.

Ex 11.4 Class 11 Maths Question 1.

$$\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ 9 } =1$$
Solution.
Given equation of hyperbola is

Ex 11.4 Class 11 Maths Question 2.

$$\frac { { y }^{ 2 } }{ 9 } -\frac { x^{ 2 } }{ 27 } =1$$
Solution.
Given equation of hyperbola is

Ex 11.4 Class 11 Maths Question 3.

9y2 – 4x2 = 36
Solution.
Given equation of hyperbola is9y2 – 4x2 = 36

Ex 11.4 Class 11 Maths Question 4.

16x2 – 9y2 = 576
Solution.
Given equation of hyperbola is16x2 – 9y2 = 576

Ex 11.4 Class 11 Maths Question 5.

5y2 – 9x2 = 36
Solution.
Given equation of hyperbola is
5y2 – 9x2 = 36

Ex 11.4 Class 11 Maths Question 6.

49y2 – 16x2 = 784
Solution.
Given equation of hyperbola is49y2 – 16x2 = 784

In each of the Exercises 7 to 15, find the equations of the hyperbola satisfying the given conditions.

Ex 11.4 Class 11 Maths Question 7.

Vertices (±2,0), foci (±3,0)
Solution.
Vertices are (±2, 0) which lie on x-axis. So the equation of hyperbola in standard form

Ex 11.4 Class 11 Maths Question 8.

Vertices (0, ±5), foci (0, ±8)
Solution.
Vertices are (0, ±5) which lie on x-axis. So the equation of hyperbola in standard form

Ex 11.4 Class 11 Maths Question 9.

Vertices (0, ±3), foci (0, ±5)
Solution.
Vertices are (0, ±3) which lie on x-axis. So the equation of hyperbola in standard form

Ex 11.4 Class 11 Maths Question 10.

Foci (±5, 0), the transverse axis is of length 8.
Solution.
Here foci are (±5, 0) which lie on x-axis. So the equation of the hyperbola in standard

Ex 11.4 Class 11 Maths Question 11.

Foci (0, ±13), the conjugate axis is of length 24.
Solution.
Here foci are (0, ±13) which lie on y-axis.
So the equation of hyperbola in standard

Ex 11.4 Class 11 Maths Question 12.

Foci ($$\pm 3\sqrt { 5 }$$,0) , the latus rectum is of length 8.
Solution.
Here foci are ($$\pm 3\sqrt { 5 }$$, 0) which lie on x-axis.
So the equation of the hyperbola in standard

Ex 11.4 Class 11 Maths Question 13.

Foci (±4, 0), the latus rectum is of length 12.
Solution.
Here foci are (±4, 0) which lie on x-axis.
So the equation of the hyperbola in standard

Ex 11.4 Class 11 Maths Question 14.

Vertices (+7, 0), e = $$\frac { 4 }{ 3 }$$
Solution.
Here vertices are (±7, 0) which lie on x-axis.
So, the equation of hyperbola in standard

Ex 11.4 Class 11 Maths Question 15.

Foci (0, $$\pm \sqrt { 10 }$$), passing through (2, 3).
Solution.
Here foci are (0, $$\pm \sqrt { 10 }$$) which lie on y-axis.
So the equation of hyperbola in standard form