NCERT Solutions For Class 11 Maths Chapter 11

NCERT Solutions Class 11 Maths Conic Sections

Ncert Solutions For Class 11 Maths Chapter 11 PDF Free Download

BYJU’S, India’s largest learning app provides free NCERT Solutions for all the classes and are available in chapter wise with an objective to help students to prepare and understand concepts in an easy manner. NCERT solutions for class 11 Maths chapter 11 conic sections provide solutions for all questions present in the class 11 maths textbooks. Those students preparing for their board exams or any other competitive exams can use this study material to have a quick review of all important equations and formulas of conic sections. The conic sections are a kind of nondegenerate curves produced by the crossings of a plane and a cone. If a plane is normal to the axis of the cone, a circle is generated. For a plane that is not normal to the axis and meets only a single nappe, the curve generated is either a parabola or an ellipse. whereas hyperbola is the curve generated by a plane meeting both nappes. Both hyperbola and ellipse are termed as central conics. You can also find the whole list of NCERT solutions for Class 11 maths, to learn different solutions from.

Due to this geometric assumption, the conics were examined by the Greeks even before its importance to the orbital inverse square law was recognised. Kepler initially noticed the ellipse shape of the orbits. Accordingly, Newton derived the pattern of orbits by utilizing calculus, by assuming that the gravitational force is inverse square of distance. Based on the energy of the orbiting object, 4 types of conic sections are possible.

Solutions for NCERT class 11 Maths chapter 11 is available in PDF format for free and are mainly prepared by the team of our subject experts to assist students in their preparations. Interested students can download the solutions for class 11 Maths chapter 11 pdf files by visiting our website at BYJU’S.

NCERT Solutions Class 11 Maths Chapter 11 Exercises

Exercise 11.1 

Q.1: A circle is given with centre (0, 3) and radius 3. Obtain the equation of the given circle.

 

Sol:

The General equation of a circle with centre (h, k) and radius r is:

(a – h)2 + (b – k)2 = r2

Given:

(h, k) = centre of a circle = (0, 3) and radius r = 3

The equation of a circle is:

(a – 0)2 + (b – 3)2 = 32

a2 + b2 – 6b + 9 = 9

a2 + b2 – 6b + 9 – 9 = 0

a2 + b2 – 6b = 0

 

 

Q.2: A circle is given with centre (-1, 2) and radius 4. Find the equation of the circle.

 

Sol:

The General equation of a circle with centre (h, k) and radius r is:

(a – h)2 + (b – k)2 = r2

Given:

Centre of a circle = (h, k) = (-1, 2) and radius (r) = 4

The equation of a circle is:

(a – (- 1))2 + (b – 2)2 = 42

(a + 1)2 + (b – 2)2 = 42

a2 + 2a + 1 + b2 – 4b + 4 = 16

a2 + 2a + b2 – 4b + 5 = 16

a2 + 2a + b2 – 4b + 5 – 16 = 0

a2 + 2a + b2 – 4b – 11 = 0

 

 

Q.3: A circle is given with centre (\(\frac{1}{3}\), \(\frac{1}{2}\)) and radius \(\frac{1}{14}\). Find the equation of the circle.

 

Sol:

The General equation of a circle with centre (h, k) and radius r is:

(a – h)2 + (b – k)2 = r2

Given:

Centre of a circle = (h, k) = (\(\frac{1}{3}\), \(\frac{1}{2}\)) and the radius (r) = 4

The equation of a circle is:

(a – \(\frac{1}{3}\))2 + (b – \(\frac{1}{2}\))2 = \(\frac{1}{14}\)2

(a – \(\frac{1}{3}\))2 + (b – \(\frac{1}{2}\))2 = \(\frac{1}{14}\)2

\(a^{2} – 2.\frac{1}{3}.a + \frac{1}{9} + b^{2} – 2.\frac{1}{2}.b + \frac{1}{4} = (\frac{1}{14})^{2} \\ a^{2} – \frac{2a}{3} + \frac{1}{9} + b^{2} – b + \frac{1}{4} = \frac{1}{196} \\ a^{2} + b^{2} – \frac{2a}{3} – b + \frac{1}{9} + \frac{1}{4} = \frac{1}{196} \\ a^{2} + b^{2} – \frac{2a}{3} – b + \frac{13}{36} = \frac{1}{196} \\ a^{2} + b^{2} – \frac{2a}{3} – b + \frac{13}{36} – \frac{1}{196} = 0 \\ a^{2} + b^{2} – \frac{2a}{3} – b + \frac{637 – 9}{1764} = 0 \\ a^{2} + b^{2} – \frac{2a}{3} – b + \frac{628}{1764} = 0 \\ a^{2} + b^{2} – \frac{2a}{3} – b + 0.35 = 0 \\ 3a^{2} + 3b^{2} – 2a – 3b + 1.068 = 0 \\\)

 

 

Q.4: A circle is given with centre (2, 2) and radius \(\sqrt{5}\). Obtain the equation of the given circle.

 

Sol:

The General equation of a circle with centre (h, k) and radius r is:

(a – h)2 + (b – k)2 = r2

Given:

Centre of a circle = (h, k) = (2, 2) and radius (r) = \(\sqrt{5}\)

The equation of a circle is:

(a – 2)2 + (b – 2)2 = \(\sqrt{5}^{2}\)

(a – 2)2 + (b – 2)2 = \(\sqrt{5}^{2}\)

a2 – 4a + 4 + b2 – 4b + 4 = 5

a2 – 4a + b2 – 4b + 8 = 5

a2 – 4a + b2 – 4b + 8 – 5 = 0

a2 – 4a + b2 – 4b + 3 = 0

a2 + b2 – 4a – 4b + 3 = 0

 

 

Q.5: A circle is given with centre (- x, – y) and radius \(\sqrt{x^{2} – y^{2}}\). Find the equation of the circle.

 

Sol:

The General equation of a circle with centre (h, k) and radius r is:

(a – h)2 + (b – k)2 = r2

Given:

Centre of a circle = (h, k) = (- x, – y) and radius ‘r’ = \(\sqrt{x^{2} – y^{2}}\)

The equation of a circle is:

(a – (- x)2 + (b – (- y)2 = \(\sqrt{x^{2} – y^{2}}^{2}\)

(a + x)2 + (b + y)2 = \(\sqrt{x^{2} – y^{2}}^{2}\)

(a + x)2 + (b + y)2 = x2 – y2

a2 + 2ax + x2 + b2 + 2by + y2 = x2 – y2

a2 + 2ax + b2 + 2by + y2 = 0

 

 

Q.6: The equation of a given circle is given as (a + 5)2 + (b – 3)2 = 36. Find the centre and radius of the circle.

 

Sol:

Given:

The equation of a given circle (a + 5)2 + (b – 3)2 = 36

(a + 5)2 + (b – 3)2 = 36

{a – (- 5)}2 + (b  – 3) = 62

Which is of the form (a – h) 2 + (b – k) 2 = r2, where h = – 5, k = 3 and r = 6.

Hence, the centre of the circle is (-5, 3), and the radius of circle is 6.

 

 

Q.7: The equation of a given circle is given as a2 + b2 – 4a – 4b + 3 = 0

Find the radius and centre of the circle.

 

Sol:

Given:

The equation of a given circle a2 + b2 – 4a – 4b + 3 = 0

a2 + b2 – 4a – 4b + 3 = 0

a2 – 4a + b2 – 4b + 3 = 0

{a2 – 2.a.2 + 22} + {b2 – 2.b.2 + 22}  – 22 – 22 + 3 = 0

{a2 – 2.a.2 + 22} + {b2 – 2.b.2 + 22}  – 8 + 3 = 0

{a2 – 2.a.2 + 22} + {b2 – 2.b.2 + 22}  = 5

(a – 2)2 + (b – 2)2 = \(\sqrt{5}^{2}\)

Which is of the form (a – h) 2 + (b – k) 2 = r2, where h = 2, k = 2 and r = \(\sqrt{5}\).

Thus, the centre of the circle is (2, 2), and the radius of circle is \(\sqrt{5}\).

 

 

Q.8: The equation of a given circle is given as a2 + b2 – 6a + 8b – 14 = 0

Find the centre and radius of the circle.

 

Sol:

Given:

 The equation of a given circle a2 + b2 – 6a + 8b – 14 = 0

a2 – 6a + b2 + 8b – 14 = 0

{a2 – 2.a.3 + 32} + {b2 + 2.b.4 + 42}  – 32 – 42 – 14 = 0

(a – 3) 2 + (b + 4) 2 – 9 – 16 – 14 = 0

(a – 3) 2 + (b + 4) 2 – 39 = 0

(a – 3) 2 + (b + 4) 2 = 39

(a – 3) 2 + (b – (- 4)) 2 = 39

Which is of the form (a – h) 2 + (b – k) 2 = r2, where h = 3, k = 4 and r = \(\sqrt{39}\).

Thus, the centre of the circle is (3, 4), and the radius of circle is \(\sqrt{39}\).

 

 

Q.9: The equation of a given circle is given as 2a2 + 2b2 – 8a = 0

Find the centre and radius of the circle.

 

Sol:

Given:

The equation of a given circle 2a2 + 2b2 – 8a = 0

2a2 + 2b2 – 8a = 0

2a2 – 8a + 2b2 = 0

2[a2 – 4a + b2] = 0

{a2 – 2.a.2 + 22} – 22 + (b – 0)2 = 0

[a2 – 2.a.2 + 22] + [b – 0]2 – 22 = 0

(a – 2)2 + (b – 0)2 = 22 = 4

Which is of the form (a – h) 2 + (b – k) 2 = r2, where h = 2, k = 0 and r = \(\sqrt{4}\) = 2.

Thus, the centre of the circle is (3, 4), and the radius of circle is 2.

 

 

Q.10: The circle passing through the points (6, 2) and (4, 3) and whose centre lies on the line 2a + y = 8. Find the equation of the circle.

 

Sol:

The equation of a required circle with centre (h, k) and radius r is given as

(a – h)2 + (b – k)2 = r2

As the circle passes through (6, 2) and (4, 3)

(6 – h)2 + (2 – k)2 = r2 . . . . . . . . . . . . . . . . (1)

(4 – h)2 + (3 – k)2 = r2 . . . . . . . . . . . . . . . . (2)

The centre (h, k) of the circle lies on the line 2a + y = 8,

2h + k = 8 . . . . . . . . . . . . . . . . . . . . . (3)

Now, On Comparing equations 1 and 2, we will get:

(6 – h)2 + (2 – k)2 = (4 – h)2 + (3 – k)2

36 – 12h + h2 + 4 – 4k + k2 = 16 – 8h + h2 + 9 – 6k + k2

36 – 12h + 4 – 4k = 16 – 8h + 9 – 6k

– 12h – 4k + 8h + 6k = 16 + 9 – 36 – 4

4h – 2k = 15 . . . . . . . . . . . . . . . . . . (4)

Now, on Solving Equations (3) and (4), we will get:

h = \(\frac{31}{8}\)

k = \(\frac{1}{4}\)

Substituting the values of h and k in equation (1), we will get:

(6 – \(\frac{31}{8}\))2 + (2 – \(\frac{1}{4}\))2 = r2

\(\\(\frac{17}{8})^{2} + (\frac{7}{4})^{2} = r^{2}\)

\(\\\Rightarrow\) \(\frac{289}{64} + \frac{49}{16} = r^{2}\)

\(\\\Rightarrow\)  \(\frac{289 + 196}{64} = r^{2}\)

\(\\\Rightarrow\) \(\frac{485}{64} = r^{2}\)

\(\Rightarrow\)  r = 7.57

The required equation is:

(a – \(\frac{31}{8}\))2 + (b – \(\frac{1}{4}\))2 = r2

(a – \(\frac{31}{8}\))2 + (b – \(\frac{1}{4}\))2 = 7.572

\((a – \frac{31}{8})^{2} + (b – \frac{1}{4})^{2} = 7.57^{2} \\\)

\(\\\Rightarrow\)   \(a^{2} – \frac{31}{4}a + (\frac{31}{8})^{2} + b^{2} – \frac{b}{2} + \frac{1}{16} = 57.30\\\)

\(\\\Rightarrow\)   \(a^{2} – \frac{31}{4}a + \frac{961}{64} + b^{2} – \frac{b}{2} + \frac{1}{16} =57.30\)

Therefore, the required equation is: 64a2– 496a + 64b2 -32b = 2702.51

 

 

Q.11: The circle passing through the points (3, 2) and (-2, 2) and whose centre lies on the line a – 3y – 13 = 0. Find the equation of the circle.

 

Sol:

The General equation of a circle with centre (h, k) and radius r is:

(a – h)2 + (b – k)2 = r2

As the circle passes through (3, 2) and (-2, 1)

(3 – h)2 + (2 – k)2 = r2 . . . . . . . . . . . . . . . . (1)

(-2 – h)2 + (2 – k)2 = r2 . . . . . . . . . . . . . . . (2)

The centre (h, k) of the circle lies on the line a – 3y – 13 = 0,

h – 3k = 13 . . . . . . . . . . . . . . . (3)

Now, on comparing equations 1 and 2, we will get:

(3 – h)2 + (2 – k)2 = (-2 – h)2 + (2 – k)2

9 – 6h + h2 = 4 + 4h + h2

9 – 6h = 4 + 4h

h = 0.5 . . . . . . . . . . . . . . . . . . (4)

Now, on solving equation (3), we will get:

k = – \(\frac{25}{6}\) = – 4.166

Now, on substituting the values of h and k in equation (1), we will get:

(3 – 0.5)2 + (2 – (- \(\frac{25}{6}\)))2 = r2

(2.5) 2 + (2 + \(\frac{25}{6}\))2 = r2

(2.5) 2 + (2 + 4.16)2 = r2

(2.5) 2 + (6.16)2 = r2

r2 = 44.1956

r = 6.65

The required equation is:

(a – h)2 + (b – k)2 = r2

(a – 0.5)2 + (b – (- 4.16))2 = r2

(a – 0.5)2 + (b + 4.16)2 = r2

a2 – a + 6.25 + b2 + 8.32b + 17.30 = 44.19

a2 – a + b2 + 8.32b = 44.19 – 6.25 – 17.30

a2 + b2 – a + 8.32b – 20.64 = 0

 

 

Q.12: The radius of the circle is 6 whose centre lies on x – axis and passes through the point (3, 2). Find the equation.

 

Sol:

Given:

The radius of the circle is 6 and passes through the point (3, 2)

The General equation of a circle with centre (h, k) and radius r is:

(a – h)2 + (b – k)2 = r2

As, the centre lies on x – axis, k = 0, the equation becomes:

(a – h) 2 + b 2 = 36

(3 – h) 2 + 22 = 36

(3 – h) 2 = 36 – 4

3 – h = \(\sqrt{32}\)

h = \(3 + 4\sqrt{2}\) or h = \(3 – 4\sqrt{2}\)

 

 

Q.13: The circle passing through (0, 0) and making intercepts a and b on the coordinate axes. Find the equation of the circle.

 

Sol:

Suppose, the equation of the required circle be (x – h) 2 + (y – k) 2 = r 2

The centre of the circle passes through (0, 0):

(0 – h) 2 + (0 – k) 2 = r 2

h 2 + k 2 = r 2

Therefore, the equation of the circle is:

(x – h) 2 + (y – k) 2 = h 2 + k 2 .

Given, the circle makes intercepts a and b on the coordinate axes.

The circle passes through points (a, 0) and (0, b).

Therefore, (a – h) 2 + (0 – k) 2 = h 2 + k 2  . . . . . . . . . . . . . . . (1)

(0 – h) 2 + (b – k) 2 = h 2 + k 2 . . . . . . . . . . . . . . . . . . (2)

Now, from equation (1), we obtain:

a 2 – 2ah + h 2 + k 2 = h 2 + k 2

a 2 – 2ah = 0

a (a – 2h) = 0

a = 0 or (a – 2h) = 0

We know, a ≠ 0; hence, (a – 2h) = 0 ⇒ h =a/2.

Now, from equation (2), we will get:

h 2 + b 2 – 2bk + k 2 = h 2 + k 2

b 2 – 2bk = 0

b (b – 2k) = 0

b = 0 or(b – 2k) = 0

However, b ≠ 0;

Therefore, (b – 2k) = 0 ⇒ k =b/2.

Thus, the equation of the required circle is:

\((x – \frac{a}{2})^{2} + (y – \frac{b}{2})^{2} = (\frac{a}{2})^{2} + (\frac{b}{2})^{2} \\ (\frac{2x – a}{2})^{2} + (\frac{2y – b}{2})^{2} = \frac{a^{2} + b^{2}}{2} \\ 4x^{2} – 4ax + a^{2} + 4y^{2} – 4bx + b^{2} = a^{2} + b^{2} \\ 4x^{2} + 4y^{2} – 4ax – 4by = 0 \\ x^{2} + y^{2} – ax – by = 0\)

 

 

Q.14: The circle with centre (3, 3) and passes through the point (6, 5), find the equation of a circle?

 

Sol:

Given:

The centre of the circle (h, k) = (3, 3).

As the circle passes through point (6, 5), the radius (r) of the circle is the distance between the points (2, 2) and (6, 5).

\(r = \sqrt{(3 – 6)^{2} + (3 – 5)^{2}} + \sqrt{(- 3)^{2} + (- 2)^{2}} = \sqrt{9 + 4} = \sqrt{13}\\\)

Therefore, the equation of the circle is:

\(\\(a – h)^{2} + (b – k)^{2} = r^{2} \\ (a – 3)^{2} + (b – 3)^{2} = \sqrt{13}^{2} \\ a^{2} – 6a + 9 + b^{2} – 6b + 9 = 13 \\ a^{2} + b^{2} – 6a – 6b + 18 – 13 = 0 \\ a^{2} + b^{2} – 6a – 6b + 5 = 0 \)

 

 

Q.15: Check whether the point (–2.0, 3.0) lies inside, outside or on the circle x 2 + y 2 = 25?

 

Sol:

Given:

The equation of the given circle is x 2 + y 2 = 25

x 2 + y 2 = 25

(x – 0)2 + (y – 0)2 = 52 , which is of the form (a – h) 2 + (b – k) 2 = r2, where h = 0, k = 0, and r = 5.

Centre = (0, 0) and radius = 5

Distance between point (– 2.0, 3.0) and centre (0, 0)

\(\sqrt{(- 2 – 0)^{2} + (3 – 0)^{2}}\\\)

=\(\sqrt{4 + 9} = \sqrt{13} = 3.60 (approx.) < 5\)

As, the distance between point (– 2.0, 3.0) and centre (0, 0) of the circle is less than the radius of the circle, point (– 2.0, 3.0) lies inside the circle.

 

 

Exercise 11.2

Q.1: For the equation y2 = 16x, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.

 

Sol:

Given:

 y2 = 16x

As the coefficient of x is positive, the given parabola has the opening on the right side.

By comparing this equation with y2 = 4ax, we get:

4a = 16 ⇒ a = 4

Therefore, focus coordinates = (a, 0) = (4, 0)

The given equation has y2, the x-axis is the axis of the parabola.

Directrix equation, x = – a = – 4

So, x + 4 = 0

Therefore, Length of the latus rectum = 4a = 4 × 4 = 16

 

 

Q.2: For the equation x2 = 8y, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.

 

Sol:

Given:

 x2 = 8y

As the coefficient of y is positive, the given parabola has the opening upwards.

By comparing this equation with x2 = 4ay, we get

4a = 8 ⇒ a = 2

Therefore, focus coordinates = (0, a) = (0, 2)

The given equation has x2, the y – axis is the axis of the parabola.

Directrix equation, y = – a = – 2

So, x + 2 = 0

Therefore, Length of the latus rectum = 4a = 4 × 2 = 8

 

 

Q.3: For the equation y2 = – 12x, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.

 

Sol:

Given:

 y2 = – 12x

As the coefficient of x is negative, the given parabola has the opening on the left side.

By comparing this equation with y2 = – 4ax, we get

4a = 12 ⇒ a = 3

Therefore, focus coordinates = (- a, 0) = (- 3, 0)

The given equation has y2, the x-axis is the axis of the parabola.

 Directrix equation, x = a = 2

So, x + 4 = 0

Therefore, Length of the latus rectum = 4a = 4 × 3 = 12

 

 

Q.4: For the equation x2 = – 20y, determine the focus coordinates,  the directrix equation, the length of the latus rectum and the axis of the parabola.

 

Sol:

Given:

 x2 = – 20y

As the coefficient of y is negative, the given parabola has the opening downwards.

By comparing this equation with x2 = – 4ay, we get:

– 4a = – 20 ⇒ a = 5

Therefore, focus coordinates = (0, a) = (0, 5)

The given equation has x2, the y – axis is the axis of the parabola.

 Directrix equation, y =  a = 5

Therefore, Length of the latus rectum = 4a = 4 × 5 = 20

 

 

Q.5: For the equation y2 = 24x, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.

 

Sol:

Given:

 y2 = 24x

As the coefficient of x is positive, the given parabola has the opening on the right side.

By comparing this equation with y2 = 4ax, we get:

4a = 24 ⇒ a = 6

Therefore, focus coordinates = (a, 0) = (6, 0)

The given equation has y2, the x-axis is the axis of the parabola.

Directrix equation, x = – a = – 6

So, x + 6 = 0

Therefore, Length of the latus rectum = 4a = 4 × 6 = 24

 

 

Q.6: For the equation x2 = – 7y, determine the focus coordinates,  the directrix equation, the length of the latus rectum and the axis of the parabola.

Sol:

Given:

 x2 = – 7y

As the coefficient of y is negative, the given parabola has the opening downwards.

By comparing this equation with x2 = – 4ay, we get:

– 4a = – 7 ⇒ a = \(\frac{7}{4}\)

Therefore, focus coordinates = (0, a) = (0, \(\frac{7}{4}\))

The given equation has x2, the y – axis is the axis of the parabola.

 Directrix equation, y = a = \(\frac{7}{4}\)

Therefore, Length of the latus rectum = 4a = 4 × \(\frac{7}{4}\) = 7

 

 

Q7. Obtain the equation of the parabola which satisfies the given conditions below:

(i) Focus coordinates = (8, 0)

(ii) Directrix (x) = – 8

 

Sol:

Given:

(i) Focus coordinates = (8, 0)

(ii) Directrix (x) = – 8

The axis of the given parabola is the x – axis as the focus is represented on the x – axis.

The required equation can be either y2 = 4ax and y2 = – 4ax

As we know, directrix (x) = – 8

As the coefficient of x is negative, the given parabola has the opening on the left side and the focus (8, 0)

Therefore, the equation of the parabola = y2 = 4ax,

Here, a = 8

Thus, the equation of the parabola = y2 = 32 x.

 

 

Q.8: Obtain the equation of the parabola which satisfies the given conditions below:

(i) Focus coordinates = (0, – 5)

(ii) Directrix (y) = 5

 

Sol:

Given:

(i) Focus coordinates = (0, – 5)

(ii) Directrix (x) = 5

The axis of the given parabola is the y – axis as the focus is represented on the y – axis.

The required equation can be either x2 = 4ay and x2 = – 4ay

As we know, directrix (x) = 5

As the coefficient of x is negative, the given parabola has the opening on the left side and the focus (0, -5)

Therefore, the equation of the parabola = x2 = – 4ay,

Here, a = 5

Thus, the equation of the parabola = x2 = – 20y

 

 

Q.9: Obtain the equation of the parabola which satisfies the given conditions below:

(i) Vertex is at origin.

(ii) Focus coordinates = (4, 0)

 

Sol:

Given:

(i) Vertex is at origin i.e., (0, 0)

(ii) Focus coordinates = (4, 0)

As the vertex of a parabola lies at origin and the focus is represented on the positive side of the x – axis, and hence, the axis of the parabola is x – axis.

The required equation is y2 = 4ax

As we know, focus is (4, 0)

Therefore, the equation of the parabola = y2 = 4ax

Here, a = 4

Thus, the equation of the parabola = y2 = 16 x

 

 

Q.10: Obtain the equation of the parabola which satisfies the given conditions below:

(i) Vertex is at origin.

(ii) Focus coordinates = (- 3, 0)

 

Sol:

Given:

(i) Vertex is at origin i.e., (0, 0)

(ii) Focus coordinates = (- 3, 0)

As the vertex of a parabola lies at origin and the focus is represented on the negative side of the x – axis, and hence, the axis of the parabola is x – axis.

The required equation is y2 = – 4ax

Therefore, the equation of the parabola = y2 = – 4ax

As we know, focus is (- 3, 0)

Here, a = 3

Thus, the equation of the parabola = y2 = – 12 x

 

 

Q.11: Obtain the equation of the parabola which satisfies the given conditions below:

(i) Vertex is at origin.

(ii) Passing through coordinates = (3, 5)

The axis of the parabola is on x – axis.

Sol:

Given:

(i) Vertex is at origin i.e., (0, 0)

(ii) Passing through coordinates = (3, 5)

As the vertex of a parabola lies at origin and the passing through coordinates = (3, 5).

The required equation is either y2 = 4ax or y2 = – 4ax

The coordinates (3, 5) is in first quadrant.

Therefore, the equation of the parabola = y2 = 4ax, whereas (3, 5) should satisfy the equation

So,

52 = 4a (3)

25 = 12 a

a =  \(\frac{25}{12}\)

The parabola’s equation is:

y2 = 4 (\(\frac{25}{12}\)) x

3 y2 = 25 x

 

 

Q.12: Obtain the equation of the parabola which satisfies the given conditions below:

(i) Vertex is at origin.

(ii) Passing through coordinates = (6, 4)

The equation is symmetric as considered with y – axis.

Sol:

Given:

(i) Vertex is at origin i.e., (0, 0)

(ii) Passing through coordinates = (6, 4)

As the vertex of a parabola lies at origin and the passing through coordinates = (6, 4).

The required equation is either x2 = 4ay or x2 = – 4ay

The coordinates (6, 4) is in first quadrant.

Therefore, the equation of the parabola = x2 = 4ay, whereas (6, 4) should satisfy the equation

So,

62 = 4a (4)

36 = 16 a

a =  \(\frac{36}{16}\)

a =  \(\frac{9}{4}\)

The parabola’s equation is:

x2 = 4(\(\frac{9}{4}\))y

x2 = 9 y

 

 

Exercise 11.3

 Q.1: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse \(\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1\).

 

Sol:

Given:

\(\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1\) . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of \(\frac{x^{2}}{25}\) > the denominator of \(\frac{y^{2}}{9}\)

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

\(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

\(c = \sqrt{m^{2} – n^{2}} = \sqrt{25 – 9} = \sqrt{16} = 4\)

We get:

m = 5, n = 3

The vertices coordinates are (m, 0) and (- m, 0) = (5, 0), (- 5, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = (4, 0) and (- 4, 0)

Length of the axis:

Major axis = 2m = 10

Minor axis = 2n = 6

Length of the latus rectum = \(\frac{2n^{2}}{m}\) = \(\frac{2.3^{2}}{5} = \frac{18}{5}\)

Eccentricity, e = \(\frac{c}{m}\) = \(\frac{4}{5}\)

 

 

Q.2: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse \(\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1\).

 

Sol:

Given:

\(\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1\) . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of \(\frac{x^{2}}{9}\) < the denominator of \(\frac{y^{2}}{16}\)

So, along y – axis consist the major axis and along the x – axis consists the minor axis.

\(\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1\) . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

\(c = \sqrt{m^{2} – n^{2}} = \sqrt{16 – 9} = \sqrt{7}\)

We get,:

m = 4, n = 3

The vertices coordinates are (0, m) and (0, – m) = (0, 4), (0, – 4)

The foci’s coordinates are (0, c) and (0, – c) = (\(\sqrt{7}\), 0) and (-\(\sqrt{7}\), 0)

Length of the axis:

Major axis = 2m = 8

Minor axis = 2n = 6

Length of the latus rectum = \(\frac{2n^{2}}{m}\) = \(\frac{2.3^{2}}{4} = \frac{9}{2}\)

Eccentricity, e = \(\frac{c}{m}\) = \(\frac{\sqrt{7}}{4}\)

 

 

Q.3: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse \(\frac{x^{2}}{25} + \frac{y^{2}}{4} = 1\).

 

Sol:

Given:

\(\frac{x^{2}}{25} + \frac{y^{2}}{4} = 1\) . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of \(\frac{x^{2}}{25}\) > the denominator of \(\frac{y^{2}}{4}\)

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

\(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\) . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

\(c = \sqrt{m^{2} – n^{2}} = \sqrt{25 – 4} = \sqrt{21}\)

We get:

m = 5, n = 2

The vertices coordinates are (m, 0) and (- m, 0) = (5, 0), (- 5, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = (\(\sqrt{21}\), 0) and (-\(\sqrt{21}\), 0)

Length of the axis:

Major axis = 2m = 10

Minor axis = 2n = 4

Length of the latus rectum = \(\frac{2n^{2}}{m}=\frac{2.2^{2}}{5} = \frac{8}{5}\)

Eccentricity, e = \(\frac{c}{m}\) = \(\frac{\sqrt{21}}{5}\)

 

 

Q.4: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse \(\frac{x^{2}}{36} + \frac{y^{2}}{121} = 1\).

 

Sol:

Given:
\(\frac{x^{2}}{36} + \frac{y^{2}}{121} = 1\) . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of \(\frac{x^{2}}{36}\) < the denominator of \(\frac{y^{2}}{121}\)

So, along y – axis consist the major axis and along the x – axis consists the minor axis.

\(\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1\) . . . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

\(c = \sqrt{m^{2} – n^{2}} = \sqrt{121 – 36} = \sqrt{85}\)

We get:

m = 11, n = 6

The vertices coordinates are (0, m) and (0, – m) = (0, 11), (0, – 11)

The foci’s coordinates are (0, c) and (0, – c) = (0, \(\sqrt{85}\)) and (0, -\(\sqrt{85}\))

Length of the axis:

Major axis = 2m = 22

Minor axis = 2n = 12

Length of the latus rectum = \(\frac{2n^{2}}{m}\) = \(\frac{2.6^{2}}{11} = \frac{72}{11}\)

Eccentricity, e = \(\frac{c}{m}\) = \(\frac{\sqrt{85}}{11}\)

 

 

Q.5: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse \(\frac{x^{2}}{64} + \frac{y^{2}}{25} = 1\).

 

Sol:

Given:

\(\frac{x^{2}}{64} + \frac{y^{2}}{25} = 1\) . . . . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of \(\frac{x^{2}}{64}\) > the denominator of \(\frac{y^{2}}{25}\)

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

\(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

\(c = \sqrt{m^{2} – n^{2}} = \sqrt{64 – 25} = \sqrt{39}\)

We get:

m = 8, n = 5

The vertices coordinates are (m, 0) and (- m, 0) = (8, 0), (- 8, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = (\(\sqrt{39}\), 0) and (-\(\sqrt{39}\), 0)

Length of the axis:

Major axis = 2m = 16

Minor axis = 2n = 10

Length of the latus rectum = \(\frac{2n^{2}}{m}\) = \(\frac{2.5^{2}}{8} = \frac{25}{4}\)

Eccentricity, e = \(\frac{c}{m}\) = \(\frac{\sqrt{39}}{8}\)

 

 

Q.6: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse \(\frac{x^{2}}{400} + \frac{y^{2}}{100} = 1\).

 

Sol:

Given:

\(\frac{x^{2}}{400} + \frac{y^{2}}{100} = 1\) . . . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of \(\frac{x^{2}}{400}\) > the denominator of \(\frac{y^{2}}{100}\)

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

\(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\) . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

\(c = \sqrt{m^{2} – n^{2}} = \sqrt{400 – 100} = \sqrt{300} = 10\sqrt{3} \)

We get:

m = 20, n = 10

The vertices coordinates are (m, 0) and (- m, 0) = (20, 0), (- 20, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = (\(10 \sqrt{3}\), 0) and (-\(10 \sqrt{3}\), 0)

Length of the axis:

Major axis = 2m = 40

Minor axis = 2n = 20

Length of the latus rectum = \(\frac{2n^{2}}{m}\) = \(\frac{2.10^{2}}{20} = 10\)

Eccentricity, e = \(\frac{c}{m}\) = \(\frac{10 \sqrt{3}}{20} = \frac{\sqrt{3}}{2}\)

 

 

Q.7: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse 49x2 + 9y2 = 441

 

Sol:

Given:

49x2 + 9y2 = 441

\(\frac{x^{2}}{9} + \frac{y^{2}}{49} = 1\)  . . . . . . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of \(\frac{x^{2}}{9}\) < the denominator of \(\frac{y^{2}}{49}\)

So, along y – axis consist the major axis and along the x – axis consists the minor axis.

\(\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1\) . . . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

\(c = \sqrt{m^{2} – n^{2}} = \sqrt{49 – 9} = \sqrt{40} = 2 \sqrt{10}\)

We get:

m = 7, n = 3

The vertices coordinates are (0, m) and (0, – m) = (0, 7), (0, – 7)

The foci’s coordinates are (0, c) and (0, – c) = (0, \(2 \sqrt{10}\)) and (0, \(2 \sqrt{10}\))

Length of the axis:

Major axis = 2m = 14

Minor axis = 2n = 6

Length of the latus rectum = \(\frac{2n^{2}}{m}\) = \(\frac{2.3^{2}}{7} = \frac{18}{7}\)

Eccentricity, e = \(\frac{c}{m}\) = \(\frac{2 \sqrt{10}}{7}\)

 

 

Q.8: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse 4x2 + 16y2 = 64

 

Sol:

Given:

4x2 + 16y2 = 64

\(\frac{x^{2}}{16} + \frac{y^{2}}{4} = 169\) . . . . . . . . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of \(\frac{x^{2}}{16}\) > the denominator of \(\frac{y^{2}}{4}\)

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

\(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

\(c=\sqrt{m^{2}-n^{2}}=\sqrt{16 – 4}=\sqrt{12}=2\sqrt{3}\)

We get:

m = 4, n = 2

The vertices coordinates are (m, 0) and (- m, 0) = (4, 0), (- 4, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = (\(2 \sqrt{3}\), 0) and (-\(2 \sqrt{3}\), 0)

Length of the axis:

Major axis = 2m = 8

Minor axis = 2n = 4

Length of the latus rectum = \(\frac{2n^{2}}{m}\) = \(\frac{2.4^{2}}{4} = 8\)

Eccentricity, e = \(\frac{c}{m}\) = \(\frac{2 \sqrt{3}}{4} = \frac{\sqrt{3}}{2}\)

 

 

Q.9: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse 4x2 + y2 = 4

 

Sol:

Given:

4x2 + y2 = 4

\(\frac{x^{2}}{1} + \frac{y^{2}}{4} = 1\) . . . . . . . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of \(\frac{x^{2}}{1}\) < the denominator of \(\frac{y^{2}}{4}\)

So, along y – axis consist the major axis and along the x – axis consists the minor axis.

\(\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1\) . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

\(c = \sqrt{m^{2} – n^{2}} = \sqrt{4 – 1} = \sqrt{3}\)

We get:

m = 2, n = 1

The vertices coordinates are (0, m) and (0, – m) = (0, 2), (0, – 2)

The foci’s coordinates are (0, c) and (0, – c) = (0, \(\sqrt{3}\)) and (0, -\( \sqrt{3}\))

Length of the axis:

Major axis = 2m = 4

Minor axis = 2n = 2

Length of the latus rectum = \(\frac{2n^{2}}{m}\) = \(\frac{2.1^{2}}{2} = 1\)

Eccentricity, e = \(\frac{c}{m}\) = \(\frac{\sqrt{3}}{2}\)

 

 

Q.10: For the given condition obtain the equation of the ellipse.

(i) Vertices (±6, 0)

(ii) Foci (±3, 0)

 

Sol:

Given:

(i) Vertices (±6, 0)

(ii) Foci (±3, 0)

The vertices are represented along x – axis.

The required equation of the ellipse is of the form \(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . . . . . (1)

We get:

m = 6 (semi major axis), c = 3

As we know:

m2 = n2 + c2

62 = n2 + 32

n2 = 62 – 32

n2 = 36 – 9 = 27

n = \(\sqrt{27} = 3\sqrt{3}\)

Hence, \(\frac{x^{2}}{36} + \frac{y^{2}}{27} = 1\)  is the equation of the ellipse.

 

 

Q.11: For the given condition obtain the equation of the ellipse.

(i) Vertices (0, ±8)

(ii) Foci ( 0, ±4)

 

Sol:

Given:

(i) Vertices (0, ±8)

(ii) Foci ( 0, ±4)

The vertices are represented along y – axis.

The required equation of the ellipse is of the form  \(\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1\) . . . . . . . . . . . . . . .(1)

We get:

m = 8 (semi major axis), c = 4

As we know:

m2 = n2 + c2

82 = n2 + 42

n2 = 82 – 42

n2 = 64 – 16 = 48

n = \(\sqrt{48} = 4\sqrt{3}\)

Hence, \(\frac{x^{2}}{48} + \frac{y^{2}}{64} = 1\) is the equation of the ellipse.

 

 

Q.12: For the given condition obtain the equation of the ellipse.

(i) Vertices (±5, 0)

(ii) Foci (±3, 0)

 

Sol:

Given:

(i) Vertices (± 5, 0)

(ii) Foci (± 3, 0)

The vertices are represented along x – axis.

The required equation of the ellipse is of the form  \(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . . . (1)

We get:

m = 5 (semi major axis), c = 3

As we know:

m2 = n2 + c2

52 = n2 + 32

n2 = 52 – 32

n2 = 25 – 9 = 16

n = \(\sqrt{16} = 4\)

Hence, \(\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1\) is the equation of the ellipse.

 

 

Q.13: For the given condition obtain the equation of the ellipse.

(i) Coordinates of major axis (±5, 0)

(ii) Coordinates of minor axis (0, ±3)

 

Sol:

Given:

(i) Coordinates of major axis (±5, 0)

(ii) Coordinates of minor axis (0, ±3)

The major axis is represented along x – axis.

The required equation of the ellipse is of the form

\(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . . . (1)

We get:

m = 5 (semi major axis), n = 3

Hence, \(\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1\) is the equation of the ellipse.

 

 

Q.14: For the given condition obtain the equation of the ellipse.

(i) Coordinates of major axis (0, ±2)

(ii) Coordinates of minor axis (±\(\sqrt{3}\), 0)

 

Sol:

Given:

(i) Coordinates of major axis (0, ±2)

(ii) Coordinates of minor axis (±\(\sqrt{3}\), 0)

The major axis is represented along y – axis.

The required equation of the ellipse is of the form  \(\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1\) . . . . . . . . . . . . . . . . . (1)

We get:

m = 2 (semi major axis), n = \(\sqrt{3}\)

Hence, \(\frac{x^{2}}{3} + \frac{y^{2}}{4} = 1\) is the equation of the ellipse.

 

 

Q.15: For the given condition obtain the equation of the ellipse.

(i) Length of major axis 30

(ii) Coordinates of foci (±4, 0)

 

Sol:

Given:

(i) Length of major axis 30

(ii) Coordinates of foci (±4, 0)

The major axis are represented along x – axis as foci is along x – axis

The required equation of the ellipse is of the form \(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . . . . . . . (1)

We get:

2m = 30 (semi major axis),

m = 15

c = 4

As we know:

 m2 = n2 + c2

152 = n2 + 42

n2 = 152 – 42

n2 = 225 – 16 = 209

n = \(\sqrt{209}\)

Hence, \(\frac{x^{2}}{225} + \frac{y^{2}}{209} = 1\) is the equation of the ellipse.

 

 

Q.16: For the given condition obtain the equation of the ellipse.

(i) Length of minor axis 26

(ii) Coordinates of foci (0, ±9)

 

Sol:

Given:

(i) Length of minor axis 26

(ii) Coordinates of foci (0, ±9)

The major axis are represented along y – axis as foci is along y – axis

The required equation of the ellipse is of the form

\(\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1\) . . . . . . . . . . . . . (1)

We get:

2n = 26 (semi major axis),

n = 13

c = 9

As we know:

m2 = n2 + c2

m2 = 132 + 92

m2 = 169 + 81

m2 = 250

m = \(\sqrt{250}\) = \(5 \sqrt{10}\)

Hence, \(\frac{x^{2}}{169} + \frac{y^{2}}{250} = 1\) is the equation of the ellipse.

 

 

Q.17: For the given condition obtain the equation of the ellipse.

(i) Coordinates of foci (±4, 0)

(ii) m = 6

 

Sol:

Given:

(i) Coordinates of foci (±4, 0)

(ii) m = 6

The major axis are represented along x – axis as foci is along x – axis

The required equation of the ellipse is of the form

\(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . . . (1)

We get:

c = 4

m = 6

As we know:

m2 = n2 + c2

62 = n2 + 42

n2 = 36 – 16

n2 = 20

n = \(\sqrt{20}\) = \(2 \sqrt{5}\)

Hence, \(\frac{x^{2}}{36} + \frac{y^{2}}{20} = 1\) is the equation of the ellipse.

 

 

Q.18: For the given condition obtain the equation of the ellipse.

(i) n = 2, c = 3 (on the x – axis)

(ii) Coordinates of centre (0, 0)

 

Sol:

Given:

(i) n = 2, c = 3 (on the x – axis)

(ii) Coordinates of centre (0, 0) i.e., centre is at origin.

The major axis are represented along x – axis as foci is along x – axis

The required equation of the ellipse is of the form:

\(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . (1)

We have:

n = 2, c = 3

As we know:

m2 = n2 + c2

m2 = 22 + 32

m2 = 4 + 9

m2 = 13

m = \(\sqrt{13}\)

Hence, \(\frac{x^{2}}{13} + \frac{y^{2}}{4} = 1\) is the equation of the ellipse.

 

 

Q.19: For the given condition obtain the equation of the ellipse.

(i) Centre is at origin and major axis is along the y – axis

(ii) It passes through the points (1, 6) and (3, 2)

 

Sol:

Given:

(i) Centre is at origin and major axis is along the y – axis

(ii) It passes through the points (1, 6) and (3, 2)

The major axis are represented along y – axis as foci is along y – axis

The required equation of the ellipse is of the form:

\(\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1\) . . . . . . . . . . . . . . . (1)

As the ellipse passes through (2, 1) and (2, 3) points, then

\(\frac{1^{2}}{n^{2}} + \frac{6^{2}}{m^{2}} = 1\)

\(\frac{3^{2}}{n^{2}} + \frac{2^{2}}{m^{2}} = 1\)

\(\frac{1}{n^{2}} + \frac{36}{m^{2}}\) = 1 . . . . . . . . . . . . . . . (a)

\(\frac{9}{n^{2}} + \frac{4}{m^{2}}\) = 1 . . . . . . . . . . . . . . . . (b)

Now, on comparing equation (a) and equation (b) we will get:

m2 = 40, n2 = 10

Hence, \(\frac{y^{2}}{40} + \frac{x^{2}}{10} = 1\) is the equation of the ellipse.

 

 

Q.20: For the given condition obtain the equation of the ellipse.

(i) Major axis is represented along x – axis

(ii) It passes through the points (6, 2) and (4, 3)

 

Sol:

Given:

(i) Major axis is represented along x – axis

(ii) It passes through the points (6, 2) and (4, 3)

The major axis are represented along y – axis as foci is along y – axis

The required equation of the ellipse is of the form:

\(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . . . . (1)

As the ellipse passes through (6, 2) and (4, 3) points, then

\(\frac{6^{2}}{m^{2}} + \frac{2^{2}}{n^{2}} = 1\)

\(\frac{4^{2}}{m^{2}} + \frac{3^{2}}{n^{2}} = 1\)

\(\frac{36}{m^{2}} + \frac{4}{n^{2}}\) = 1 . . . . . . . . . . . . . . . . (a)

\(\frac{16}{m^{2}} + \frac{9}{n^{2}}\) = 1 . . . . . . . . . . . . . . . . (b)

Now, on comparing equation (a) and equation (b) we will get:

m2 = 52 and n2 = 13

Hence, \(\frac{y^{2}}{13} + \frac{x^{2}}{52} = 1\) is the equation of the ellipse.

 

 

Exercise 11.4

Q.1: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola \(\frac{x^{2}}{25} – \frac{y^{2}}{16} = 1\).

 

Sol:

Given:

\(\frac{x^{2}}{25} – \frac{y^{2}}{16} = 1\) . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is \(\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . . . . . .(2)

Now, on comparing equation (1) and equation (2) we will get:

\(c = \sqrt{m^{2} + n^{2}} = \sqrt{25 + 16} = \sqrt{41}\)

We get:

m = 5 and n = 4

The vertices coordinates are (m, 0) and (- m, 0) = (5, 0), (- 5, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = (\(\sqrt{41}\), 0) and (-\(\sqrt{41}\), 0)

Length of the latus rectum = \(\frac{2n^{2}}{m}\) = \(\frac{2.4^{2}}{5} = \frac{32}{5}\)

Eccentricity, e = \(\frac{c}{m}\) = \(\frac{\sqrt{41}}{5}\)

 

 

Q.2: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola \(\frac{x^{2}}{1} – \frac{y^{2}}{8} = 1\).

 

Sol:

Given:

\(\frac{x^{2}}{1} – \frac{y^{2}}{8} = 1\) . . . . . . . . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is \(\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . . . . . . . . . . .(2)

Now, on comparing equation (1) and equation (2) we will get:

m = 1and  n = \(\sqrt{8}\)

As we know:

\(c = \sqrt{m^{2} + n^{2}} = \sqrt{1 + 8} = \sqrt{9}\) = 3

The vertices coordinates are (0, m) and (0, – m) = (0, 1), (0, – 1)

The foci’s coordinates are (0, c) and (0, – c) = (0, 3), (0, – 3)

Length of the latus rectum = \(\frac{2n^{2}}{m}\) = \(\frac{16}{1} = 16\)

Eccentricity, e = \(\frac{c}{m}\) = \(\frac{\sqrt{3}}{1}\) = 3

 

 

Q.3: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 25 y2 – 4 x2 = 100

 

Sol:

Given:

25 y2 – 4 x2 = 100

\(\frac{y^{2}}{4} – \frac{x^{2}}{25} = 1\) . . . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is \(\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . . .(2)

Now, on comparing equation (1) and equation (2) we will get:

m = 2 and n = 5.

\(c = \sqrt{m^{2} + n^{2}} = \sqrt{4 + 25} = \sqrt{29}\)

The vertices coordinates are (0, m) and (0, – m) = (0, 2) and (0, – 2)

The foci’s coordinates are (0, c) and (0, – c) = (0, \(\sqrt{29}\)) and (0, -\(\sqrt{29}\))

Length of the latus rectum = \(\frac{2.n^{2}}{m}\) = \(\frac{2.5^{2}}{2} = \frac{50}{2}\) = 25

Eccentricity, e = \(\frac{c}{m}\) = \(\frac{\sqrt{29}}{2}\)

 

 

Q.4: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 9 x2 – 4 y2 = 36

 

Sol:

Given:

9 x2 – 4 y2 = 36

\(\frac{x^{2}}{4} – \frac{y^{2}}{9} = 1\) . . . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is \(\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

m = 2 and n = 3.

\(c = \sqrt{m^{2} + n^{2}} = \sqrt{4 + 3} = \sqrt{13}\)

The vertices coordinates are (m, 0) and (- m, 0) = (2, 0), (- 2, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = (\(\sqrt{13}\), 0) and (-\(\sqrt{13}\), 0)

Length of the latus rectum = \(\frac{2n^{2}}{m}\) = \(\frac{2.3^{2}}{2} = 9\)

Eccentricity, e = \(\frac{c}{m}\) = \(\frac{\sqrt{13}}{2}\)

 

 

Q.5: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 6 y2 – 16 x2 = 96

 

Sol:

Given:

6 y2 – 16 x2 = 96

\(\frac{y^{2}}{16} – \frac{x^{2}}{6} = 1\) . . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is \(\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . . . . . .(2)

Now, on comparing equation (1) and equation (2) we will get:

m = 4 and n = \(\sqrt{6}\)

\(c = \sqrt{m^{2} + n^{2}} = \sqrt{4 + 6} = \sqrt{10}\)

The vertices coordinates are (0, m) and (0, – m) = (0, 4) and (0, – 4)

The foci’s coordinates are (0, c) and (0, – c) = (0, \(\sqrt{10}\)) and (0, -\(\sqrt{10}\))

Length of the latus rectum = \(\frac{2.n^{2}}{m}\) = \(\frac{2. \sqrt{6}^{2}}{4} = \frac{12}{4}\) = 3

Eccentricity, e = \(\frac{c}{m}\) = \(\frac{\sqrt{10}}{4}\)

 

 

Q.6: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 64 y2 – 9 x2 = 96

 

Sol:

Given:

6 y2 – 16 x2 = 96

\(\frac{y^{2}}{16} – \frac{x^{2}}{6} = 1\) . . . . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is \(\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

m = 4 and n = \(\sqrt{6}\)

\(c = \sqrt{m^{2} + n^{2}} = \sqrt{4 + 6} = \sqrt{10}\)

The vertices coordinates are (0, m) and (0, – m) = (0, 4) and (0, – 4)

The foci’s coordinates are (0, c) and (0, – c) = (0, \(\sqrt{10}\)) and (0, -\(\sqrt{10}\))

Length of the latus rectum = \(\frac{2.n^{2}}{m}\) = \(\frac{2. \sqrt{6}^{2}}{4} = \frac{12}{4}\) = 3

Eccentricity, e = \(\frac{c}{m}\) = \(\frac{\sqrt{10}}{4}\)

 

 

Q.7: For the given condition obtain the equation of the hyperbola.

(i) Coordinates of the vertices (±3, 0)

(ii) Coordinates of the foci (±4, 0)

 

Sol:

Given:

(i) Coordinates of the vertices (±3, 0)

(ii) Coordinates of the foci (±4, 0)

The specific equation for hyperbola is \(\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . . . . (1)

As, coordinates of the vertices (±3, 0), m = 3

And, coordinates of the foci (±4, 0), c = 4

We know that:

c2 = m2 + n2

42 = 32 + n2

n2 = 16 – 9 = 5

\(\frac{x^{2}}{9} – \frac{y^{2}}{5} = 1\) is the equation of the hyperbola.

 

 

Q.8: For the given condition obtain the equation of the hyperbola.

 (i) Coordinates of the vertices (0, ±7)

(ii) Coordinates of the foci (0, ±9)

 

Sol:

Given:

(i) Coordinates of the vertices (0, ±7)

(ii) Coordinates of the foci (0, ±9)

The specific equation for hyperbola is \(\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . . . . . . . (1)

As, coordinates of the vertices (0, ±7), m = 7

And, coordinates of the foci (0, ±9), c = 9

We know that:

c2 = m2 + n2

92 = 72 + n2

n2 = 81 – 49 = 32

\(\frac{y^{2}}{49} – \frac{x^{2}}{32} = 1\) is the equation of the hyperbola.

 

 

Q.9: For the given condition obtain the equation of the hyperbola.

 (i) Coordinates of the vertices (0, ±8)

(ii) Coordinates of the foci (0, ±11)

 

Sol:

Given:

(i) Coordinates of the vertices (0, ±8)

(ii) Coordinates of the foci (0, ±11)

The specific equation for hyperbola is \(\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . . . . . . . . . . (1)

As, coordinates of the vertices (0, ±8), m = 8

And, coordinates of the foci (0, ±11), c = 11

We know that:

c2 = m2 + n2

112 = 82 + n2

n2 = 121 – 64 = 57

\(\frac{y^{2}}{64} – \frac{x^{2}}{57} = 1\) is the equation of the hyperbola.

 

 

Q.10: For the given condition obtain the equation of the hyperbola.

 (i) Coordinates of the foci (±7, 0),

(ii) The length of the transverse axis is 12

 

Sol:

Given:

(i) Coordinates of the foci (±7, 0),

(ii) The length of the transverse axis is 12

The specific equation for hyperbola is \(\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . . . . . . (1)

And, coordinates of the foci (±7, 0), c = 7

As, the transverse axis is of length 12,

2 m = 12

m = 6

We know that:

c2 = m2 + n2

72 = 62 + n2

n2 = 49 – 36 = 13

\(\frac{x^{2}}{36} – \frac{y^{2}}{13} = 1\) is the equation of the hyperbola.

 

 

Q.11: For the given condition obtain the equation of the hyperbola.

(i) Coordinates of the foci (0, ±8)

(ii) The length of the conjugate axis is 16

 

Sol:

Given:

(i) Coordinates of the foci (0, ±9)

(ii) The length of the conjugate axis is 16

The specific equation for hyperbola is \(\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . . . . (1)

And, coordinates of the foci (0, ±9), c = 9

As, the conjugate axis is of length 16,

2 m = 16

m = 8

Therefore, m2 = 64

We know that:

c2 = m2 + n2

92 = 82 + n2

n2 = 81 – 64 = 17

\(\frac{y^{2}}{64} – \frac{x^{2}}{17} = 1\) is the equation of the hyperbola.

 

 

Q.12: For the given condition obtain the equation of the hyperbola.

(i) Coordinates of the foci (±\(2\sqrt{3}\), 0 )

(ii) The length of the latus rectum is 8

 

Sol:

Given:

(i) Coordinates of the foci (±\(2\sqrt{3}\), 0 )

(ii) The length of the latus rectum is 8

The specific equation for hyperbola is \(\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . (1)

And, the coordinates of the foci (±\(2\sqrt{3}\), 0 ), c = \(2\sqrt{3}\)

As, the latus rectum is of length 8

\(\frac{2n^{2}}{m} = 8 \\ 2n^{2} = 8 m \\ n^{2} = 4 m\)

We know that:

m2 + n2 = c2

m2 + 4m = \((2\sqrt{3})^{2}\)

m2 + 4m = 12

m2 + 4m – 12 = 0

m2 + 6m – 2m – 12 = 0

m (m + 6) – 2 (m + 6) = 0

m = – 6 and m = 2

m is non – negative so, m = 2 and n2 = 8 [Since, n2 = 12 – 22 = 8]

\(\frac{x^{2}}{4} – \frac{y^{2}}{8} = 1\) is the equation of the hyperbola.

 

 

Q.13: For the given condition obtain the equation of the hyperbola.

(i) Coordinates of the foci (±\(2\sqrt{6}\), 0)

(ii) The length of the latus rectum is 10

 

Sol:

Given:

(i) Coordinates of the foci (±\(2\sqrt{3}\), 0 )

(ii) The length of the latus rectum is 10

The specific equation for hyperbola is \(\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . . . . . (1)

And, coordinates of the foci (±\(2\sqrt{6}\), 0 ), c = \(2\sqrt{6}\)

As, the latus rectum is of length 10

Therefore, \(\frac{2n^{2}}{m} = 10 \\ 2n^{2} = 10 m \\ n^{2} = 5m\)

We know that:

 m2 + n2 = c2

m2 + 5 m = \((2\sqrt{6})^{2}\)

m2 + 5 m = 24

m2 + 5 m – 24 = 0

m2 + 8 m – 3 m – 24 = 0

m (m + 8) – 3 (m + 8) = 0

m = – 8 and m = 3

m is non – negative so, m = 3 and n2 = 5m = 15

\(\frac{x^{2}}{9} – \frac{y^{2}}{15} = 1\) is the equation of the hyperbola.

 

 

Q.14: For the given condition obtain the equation of the hyperbola.

(i) Coordinates of the vertices (±9, 0)

(ii) Eccentricity (e) = \(\frac{9}{4}\)

 

Sol:

Given:

 (i) Coordinates of the vertices (±9, 0)

(ii) Eccentricity (e) = \(\frac{9}{4}\)

The specific equation for hyperbola is \(\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . . . . . (1)

And, Coordinates of the vertices (±9, 0), m = 9

As, the eccentricity (e) = \(\frac{9}{4}\),

\(\frac{c}{m} = \frac{9}{4}\)

\(\frac{c}{9} = \frac{9}{4}\)

c = \(\frac{81}{4}\)

We know that:

 m2 + n2 = c2

92 + n2 = \((\frac{81}{4})^{2}\)

n2 = \((\frac{81}{4})^{2}\) – 92

n2 = \(\frac{6561}{16} – 81\)

n2 = \(\frac{6561 – 1296}{16} = \frac{5265}{16}\)

\(\frac{x^{2}}{81} – \frac{16 y^{2}}{5265} = 1\) is the equation of the hyperbola.

 

 

Q.15: For the given condition obtain the equation of the hyperbola.

(i) Coordinates of the foci (0, ± \(\sqrt{10}\))

(ii) It passes through (2, 3)

 

Sol:

Given:

(i) Coordinates of the foci (0, ±\(\sqrt{10}\))

(ii) It passes through (2, 3)

The specific equation for hyperbola is \(\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1\) . . . . . . . . . . . . (1)

And, coordinates of the foci (0, ±\(\sqrt{10}\)), c = \(\sqrt{10}\)

We know that:

 m2 + n2 = c2

n2 = c2 – m2

n2 = 10 – m2 . . . . . . . . . . . . (2)

As the hyperbola passes through (2, 3)

\(\frac{3^{2}}{m^{2}} – \frac{2^{2}}{n^{2}} = 1\)

\(\frac{9}{m^{2}} – \frac{4}{n^{2}} = 1\) . . . . . . . . . . . (3)

Putting Equation (2) in equation (3), we get:

\(\frac{9}{m^{2}}-\frac{4}{10 – m^{2}}= 1\)

9 (10 – m2) – 4m2 = m2 (10 – m2)

m4 – 23m2 + 90 = 0

m4 – 18m2 – 5m2 + 90 = 0

(m2 – 5) (m2 – 18) = 0

m2 = 5 or 18

As in hyperbola:

m2 < c2

m2 = 5

n2 = 10 – 5 [From Equation (2)] = 5

\(\frac{y^{2}}{5} – \frac{x^{2}}{5} = 1\) is the equation of the hyperbola.