## MP Board Class 11th Maths Important Questions Chapter 11 Conic Sections

### Conic Sections Important Questions

**Conic Sections Objective Type Questions**

(A) Choose the correct option :

Question 1.

Coordinates of the focus of the parabola y = 2x^{2} + x are:

(a) (0, 0)

(b) (\(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 4 }\))

(c) (- \(\frac { 1 }{ 4 }\), 0)

(d) ( – \(\frac { 1 }{ 4 }\), \(\frac { 1 }{ 8 }\))

Answer:

(c) (- \(\frac { 1 }{ 4 }\), 0)

Question 2.

In a ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 a > b, the relation between a, b an eccentricity e is:

(a) b^{2} = a^{2}(1 – e^{2})

(b) b^{2} = a^{2}(e^{2} – 1)

(c) a^{2} = b^{2}(1 – e^{2})

(d) a^{2} = b^{2}(e^{2} – 1)

Answer:

(a) b^{2} = a^{2}(1 – e^{2})

Question 3.

The length of latus rectum of ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1, represent a circle then its eccentricity will be:

(a) \(\frac { { 2a }^{ 2 } }{ b }\)

(b) \(\frac { { 2b }^{ 2 } }{ a }\)

(c) \(\frac { { a }^{ 2 } }{ b }\)

(d) \(\frac { { b }^{ 2 } }{ a }\)

Answer:

(b) \(\frac { { 2b }^{ 2 } }{ a }\)

Question 4.

The eccentricity of the parabola is:

(a) Less than 1

(b) Greater than 1

(c) 0

(d) 1

Answer:

(d) 1

Question 5.

The eccentricity of the ellipse is:

(a) Less than 1

(b) Greater than 1

(c) 0

(d) 1

Answer:

(a) Less than 1

Question 6.

The eccentricity of the hyperbola is:

(a) Less than 1

(b) Greater than 1

(c) 0

(d) 1

Answer:

(b) Greater than 1

Question 7.

In a ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1, represent a circle then its eccentricity will be:

(a) Less than 1

(b) Greater than 1

(c) 0

(d) 1

Answer:

(c) 0

Question 8.

The sum of focal distances from any point on the ellipse is:

(a) Equal to major axis

(b) Equal to minor axis

(c) The distance between two foci

(d) Equal to latus rectum.

Answer:

(a) Equal to major axis

Question 9.

The differecne of the focal distances from any point on the hyperbola is:

(a) Equal to its conjugate axis

(b) Equal to its transverse axis

(c) The distance between two foci

(d) Equal to its latus rectum.

Answer:

(b) Equal to its transverse axis

Question 10.

The value of the eccentricity of ellipse 25x^{2} + 16y^{2} = 400 is:

(a) \(\frac { 3 }{ 5 }\)

(b) \(\frac { 1 }{ 3 }\)

(c) \(\frac { 2 }{ 5 }\)

(d) \(\frac { 1 }{ 5}\)

Answer:

(a) \(\frac { 3 }{ 5 }\)

Question 11.

Equation ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 represent a circle if:

(a) a = b, c = 0

(b) f = g, h = 0

(c) a = b, h = 0

(d) f = g, c = 0

Answer:

(a) a = b, c = 0

Question 12.

Area of triangle whose centre (1,2) and which is passes through the point (4,6) will be:

(a) 5π

(b) 10π

(c) 25π

(d) 25π^{2}

Answer:

(c) 25π

Question 13.

The circle passing through (1, – 2) and touching the X – axis at (3,0), also passes through the point:

(a) (2, – 5)

(b) (5, – 2)

(c) (- 2, 5)

(d) (- 5, 2)

Answer:

(a) (2, – 5)

Question 14.

The length of the diameter of the circle which touches the X – axis at the point (1,0) and passes through the point (2,3) is:

(a) \(\frac { 10 }{ 3 }\)

(b) \(\frac { 3 }{ 5 }\)

(c) \(\frac { 6 }{ 5 }\)

(d) \(\frac { 5 }{ 3 }\)

Answer:

(a) \(\frac { 10 }{ 3 }\)

Question 15.

Eccentricity of the hyperbola 3x^{2} – y^{2} = 4 :

(a) 1

(b) 2

(c) – 2

(d) \(\sqrt {2}\)

Answer:

(b) 2

(B) Match the following :

Answer:

- (c)
- (e)
- (b)
- (a)
- (d)
- (i)
- (h)
- (f)
- (j)
- (g)

(C) Fill in the blanks :

- The length of the latus rectum of the parabola y
^{2}= 4ax is …………… - The centre of the ellipse \(\frac { { (x-1) }^{ 2 } }{ 9 } +\frac { { (y-2) }^{ 2 } }{ 4 }\) = 1 will be ……………
- The vertex of the parabola (y – 2)
^{2}= 4a(x -1) is …………… - The lines \(\frac { x }{ a }\) – \(\frac { y }{ b }\) = m and \(\frac { x }{ a }\) + \(\frac { y }{ b }\) = \(\frac { 1 }{ m }\) meets always at ……………
- If an ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1, a > b and its eccentricity is e, then the foci will be
- Standard form of equation of parabola is ……………
- Parametric equation of a circle x
^{2}+ y^{2}= 4 is …………… - A line y = x + a\(\sqrt {2}\) touches the circle x
^{2}+ y^{2}= a^{2}point …………… - A line y = mx + c touches the circle x
^{2}+ y^{2}= a^{2}if c = …………… - Vertex of the parabola 3y
^{2}+ 6y – 4x + 11 = 0 is …………… - Equation 2x
^{2}+ 2y^{2}– 12x – 16y + 4 = 0 represent a point circle if k = …………… - Radius of circle 3x
^{2}+ 3y^{2}– 5x – 6y + 4 = 0 is ……………

Answer:

- 4a
- (1, 2)
- (1, 2)
- Hyperbola
- (± ae, 0)
- y
^{2}= 4ax - x = 2cosθ
- (- \(\frac { a }{ \sqrt { 2 } }\), \(\frac { a }{ \sqrt { 2 } }\) )
- ±a\(\sqrt { 1+{ m }^{ 2 } }\)
- (5, 1)
- 50, 12
- \(\sqrt {61}\)

(D) Write true / false :

- Conic section is a locus of the point whose the ratio between the distance from the fixed point and distance from the fixed line, this ratio is called eccentricity of the conic section.
- The ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 has two directrics, the equation of directrics are x = ± \(\frac { a }{ e }\); Where a > b and y = ± \(\frac { b }{ e }\) ; where b > a.
- The foci of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 are (0, ± be) where a < b.
- A circle drawn by taking major axis of the ellipse as diameter is called auxiliary circle of the ellipse.
- The locus of intersection point of the lines bx + ay = abt and bx – ay = \(\frac { ab}{ t }\) will be a ellipse.
- The focus of a parabola x
^{2}= – 16y will be (0, – 4). - Equation x
^{2}+ y^{2}– 6x + 8y + 50 = 0 represent a circle. - Circle x
^{2}+ y^{2}= 9 and x^{2}+ y^{2}+ 8y + c = 0 touches externally if c = 15 . - Eccentricity of hyperbola is 1.
- Minimum distance between line y – x = 1 and curve x = y
^{2}is \(\frac { 3\sqrt { 2 } }{ 8 }\)

Answer:

- True
- True
- True
- True
- False
- True
- False
- True
- False
- True

(E) Write answer in one word / sentence :

- If the circle x
^{2}+ y^{2}+ 2ax + 8y +16 = 0, touches X – axis, then the value of α will be. - Coordinate of focus of parabola x
^{2}= – 10y will be. - Write the equation of a circle whose centre is (2,2) and passes through the point (4, 5).
- The centre of a circle is (5, 7) and touches Y – axis, then its radius will be.
- If the radius of a circle x
^{2}+ y^{2}– 6x + ky – 25 = 0 is \(\sqrt {38}\) the value of k will be. - Vertex of the parabola y = x
^{2}– 2x + 3 will be. - Equation of a parabola whose vertex (0, 0) and focus (0, 3) will be.
- Length of major axis of ellipse 9x
^{2}+ 16y^{2}= 144 will be. - Eccentricity of an ellipse whose latus rectum in half of its minor axis will be.
- Equation of hyperbola whose one focus in (4, 0 ) and corresponding equation of directrix x = 1 will be.

Answer:

- ± 4
- (0, \(\frac { – 5 }{ 2 }\))
- x
^{2}+ y^{2}– 4x – 4y – 5 = 0 - 7, 5
- ± 4
- (1, 2)
- x
^{2}= 12y - 6
- \(\frac { \sqrt { 3 } }{ 2 }\)
- \(\frac { x^{ 2 } }{ 4 } -\frac { y^{ 2 } }{ 12 } \) = 1

**Conic Sections Long Answer Type Questions**

Question 1.

Find the equation of circle which touches the X – axis at a distance of 4 units in the negative direction and makes intercept of 6 units on positive direction of Y – axis.

Solution:

Here OA = CM = 4, BD = 6.

Length of perpendicular drawn from centre C on BD.

Then, BM = MD = 3

In right angled ∆ CMB,

CB^{2} = CM^{2} + BM^{2}

= 4^{2} + 3^{2}

= 16 + 9 =25

⇒ CB = 5

∴ CA = Radius of circle = CB = 5

∴ Centre of circle (- 4, 5) and radius = 5

Hence, equation of circle :

(x + 4)^{2} + (y – 5)^{2} = 5^{2}

⇒ x^{2} + 8x + 16 + y^{2} – 10y + 25 = 25

⇒ x^{2} + y^{2} + 8x – 10y + 16 = 0

Question 2.

Find the equation of circle which touches Y – axis at a distance of 4 units and makes intercept of 6 units on Y – axis?

Solution:

Given : OP = 4, AB = 6, PC = AC = radius.

CM ⊥ AB ∴ AM = BM = \(\frac { 6 }{ 2 }\) = 3

OP = CM = 4

In right angled ∆ AMC,

AC^{2} = AM^{2} + CM^{2}

= (3)^{2} + (4)^{2} = 9 + 16 = 25

∴ AC = 5

From figure PC – OM= 5 = radius

Centre of circle is (5, 4) and radius = 5.

Hence, required equation of circle :

(x – 5)^{2} + (y – 4)^{2} = (5)^{2}

x^{2} – 10x + 25 + y^{2} – 8y + 16 = 25

x^{2} + y^{2} – 10x – 8y + 16 = 0.

Question 3.

ABCD is a square. Supposing AB and AD as the coordinate axes. Find the equation of the circle circumscribing the square if each side of square is of length l.

Solution:

Taking AB and AD as X – axis and Y – axis respectively

Given : AB = BC = CD = DA = 1

M is mid point of AB.

N is mid point of AD.

AM = \(\frac { l }{ 2 }\), AN = \(\frac { l }{ 2 }\) = OM

In ∆OAM,

OA^{2} = AM^{2} + OM^{2}

= \(\frac { l }{ 2 }\)^{2} + \(\frac { l }{ 2 }\)^{2}

= \(\frac{l^{2}}{4}+\frac{l^{2}}{4} = \frac{l^{2}}{2}\)

∴ Radius = OA = \(\frac{l}{\sqrt{2}}\)

Centre of circle (AM, OM) = ( \(\frac { l }{ 2 }\), \(\frac { l }{ 2 }\) )

Required equation of circle is :

(x – \(\frac { l }{ 2 }\) )^{2} + (y – \(\frac { l }{ 2 }\) )^{2} = \(\frac{l^{2}}{2}\)

Centre of circle (AM, OM) = (\(\frac { l }{ 2 }\), \(\frac { l }{ 2 }\))

Required equation of circle is :

(x – \(\frac { l }{ 2 }\))^{2} + (y – \(\frac { l }{ 2 }\))^{2} = \(\frac{l^{2}}{2}\)

⇒ x^{2} – lx + \(\frac{l^{2}}{4}\) + y^{2} – ly + \(\frac{l^{2}}{4}\) = \(\frac{l^{2}}{2}\)

⇒ x^{2} + y^{2} – l(x + y) = 0

⇒ x^{2} + y^{2} = l(x + y)

Question 4.

Find the equation of the circle passing through the points (4, 1) and (6, 5). Whose centre lies on line 4x + y = 16. (NCERT)

Solution:

Let the equation of circle be

x^{2} + y^{2} + 2gx + 2 fy + c = 0 …. (1)

It passes through points (4, 1) and (6, 5).

∴ 8g + 2f + c + 17 = 0 …. (2)

and 12g + 10f + c + 61 = 0 …. (3)

Centre of circle (1) is (- g, – f) which lies on line 4x + y = 16.

∴ – 4g – f – 16 = 0

⇒ 4g + f + 16 – 0 …. (4)

Subtracting equation (2) from equation (3), we get

4g + 8f + 44 = 0

⇒ g + 2f + 11 = 0 …. (5)

On solving equation (4) and (5), g = – 3, f = – 4

Put g = – 3 and f = – 4 in equation (2),

– 24 – 8 + C + 17 = 0

⇒ c = 15

Put values of g, f and c in equation (1), then required equation of circle is :

x^{2} + y^{2} – 6x – 8y + 15 = 0.

Question 5.

Find the equation of the circle which passes through the points (2, 3) and (- 1, 1) whose centre lies on line x – 3y – 11 = 0. (NCERT)

Solution:

Let the equation of circle is :

x^{2} + y^{2} + 2gx + 2fy + c = 0 …. (1)

∵Points (2, 3) and (- 1, 1) lies on equation (1),

∴ (2)^{2} + (3)^{2} + 2g(2) + 2f(3) + c = 0

⇒ 4 + 9 + 4g + 6f + c + 13 = 0

4g + 6f + c + 13 = 0 …. (2)

and (-1)^{2} + (l)^{2} – 2g + 2f + c = 0

⇒ 1 + 1 – 2g + 2f + c = 0

⇒ – 2g + 2f + c + 2 = 0 …. (3)

Putting the value of g, f and c in equation (1), then required equation of circle will be :

x^{2} + y^{2} + 2(\(\frac { -7 }{ 2 }\))x + 2(\(\frac { 5 }{ 2 }\))y + c = 0 …. (1)

x^{2} + y^{2} – 7x + 5y -14 = 0.

Question 6.

Find the equation of circle whose radius is 5, centre is on Y – axis and which passes through point (2, 3).

Solution:

Centre of circle is on X – axis, so k = 0.

Let the equation of circle be :

(x – h)^{2} + (y – k)^{2} = a^{2}

Here a = 5

(x – h)^{2} + (y – 0)^{2} = (5)^{2}

(x – h)^{2} + y^{2} = 25

Circle (1) passes through point (2, 3),

∴ (2 – h)^{2} + (3)^{2} = 25

⇒ (2 – h)^{2} = 25 – 9 = 16 = (4)^{2}

⇒ 2 – h = ± 4

Question 7.

y = mx is a chord of the circle whose radius is ‘a’ and its diameter is X – axis. Origin is one of the limiting points of the chord. Show that the equation to a circle whose diameter is the given chord is given by the equation (1 + m^{2}) (x^{2} + y^{2} ) – 2a (x + my) = 0. Solution:

Equation of circle whose radius is a and centre is (a, 0) will be

(x – a)^{2} + y^{2} = a^{2}

⇒ x^{2} – 2ax + y^{2} + a^{2} = a^{2}

⇒ x^{2} – 2ax + y^{2} = 0 …. (1)

Equation of given line is :

y = mx …. (2)

Now, equation of circle passing through the intersection of eqns. (1) and (2) will be :

x^{2} + y^{2} – 2ax + λ(y – mx) = 0 … (3)

Centre of co – ordinate of circle (3) are (\(\frac { λm + 2a }{ 2 }\), \(\frac { λ }{ 2 }\))

∵ Centre lies on line y = mx.

∴ – \(\frac { λ }{ 2 }\) = m\(\frac { λm + 2a }{ 2 }\)

⇒ λ = \(\frac{-2 a m}{1+m^{2}}\)

Put the value of λ in Equation (3),

x^{2} + y^{2} – 2ax + \(\frac{-2 a m}{1+m^{2}}\) (y – mx) = 0

⇒ (l + m^{2})(x^{2} + y^{2}) = 2ax + 2am^{2}x + 2amy – 2am^{2}x

⇒ (l + m^{2})(x^{2} + y^{2}) = 2a(x + my)

⇒ (l + m^{2})(x^{2} + y^{2}) – 2a(x + my) = 0

Which is required equation of circle.

Question 8.

If the straight line x cos α + y sin α = p cuts a circle x^{2} + y^{2} = a^{2} in two points M and N, then show that the equation of the circle whose diameter is MN will be x^{2} + y^{2} – a^{2} = 2p(x cos α + y sin α – p).

Solution:

Given : Equation of line is :

x cos α + y sin α = p …. (1)

and Equation of circle is

x^{2} + y^{2} = a^{2} …. (2)

Now, equation of circle passing through the intersection of line (1) and circle (2) at points M and N is :

x^{2} + y^{2} – a^{2} + λ(x cos α + y sin α – p) = 0 …. (3)

If MN is diameter of above circle then centre is :

(- \(\frac { λ }{ 2 }\)cos α, – \(\frac { λ }{ 2 }\)sin α)

Which is lies on line x cos α + y sin α = p.

– ( \(\frac { λ }{ 2 }\)cos α )cos α + (- \(\frac { λ }{ 2 }\)sin α)sin α = p

⇒ – \(\frac { λ }{ 2 }\)[cos^{2 }α + sin^{2 }α] = p

⇒ λ = – 2p

Put the value of λ in equation (3), then required equation of circle is

x^{2} + y^{2} – a^{2} – 2p(x cos α + y sin α – p) = 0

⇒ x^{2} + y^{2} – a^{2} = 2p(x cos α + y sin α – p)

Question 9.

Find the following equation of parabola : (i) co – ordinates of focus, (ii) axis, (iii) equation of directrix, (iv) length of Iatus rectum. (NCERT)

(A) y^{2} = 12x

Solution:

Equation of parabola : y^{2} = 12x

Comparing with y^{2} = 4 ax,

4a = 12 ⇒ a = 3

∴Co – ordinates of focus (a, 0) = (3, 0).

Axis of parabola = X – axis.

Equation of directrix is x = – a ⇒ x = – 3.

Length of latus rectum = 4a = 4 x 3 = 12.

(B) x^{2} = 6y.

Solution:

Equation of parabola: x^{2} = 6y

Comparing with x^{2} = 4ay

4a = 6 a ⇒ 3/2

∴ Co – ordinate of focus (0, a) = (0, 3/2).

Axis of parabola = Y – axis.

Equation of directrix is y = – a ⇒ y = – 3/2.

(C) y^{2} = – 8x

Solution:

Equation of parabola : y^{2} = – 8x

Comparing with y^{2} = – 4ax

– 4a – = – 8 ⇒ a = 2

∴ Co – ordinates of focus (- a, 0) = (- 2, 0).

Axis of parabola = X – axis.

Equation of directrix is x = a ⇒ x = 2.

Length of latus rectum 4a = 4 x 2 = 8.

(D) x^{2} = – 16y

Solution:

Equation of parabola : x^{2} = – 16y

Comparing with x^{2} = – 4ay

– 4a = – 16 ⇒ a = 4

∴ Co – ordinate of focus (0, – a) = (0, – 4)

Axis of parabola = Y – axis

Equation of directrix is y = a ⇒ y = 4

Length of latus rectum = 4a = 16.

Question 10.

An equilateral triangle inscribed in the parabola y^{2} = 4ax, where one vertex is at the vertex of parabola. Find the length of the side of triangle. (NCERT)

Solution:

Let the equation of parabola is y^{2} = 4ax.

Let APQ be the equilateral triangle whose vertex A(0, 0), P(h, k) and Q(h, – k).

AP^{2} = (h – 0)^{2} + (k – 0)^{2}

= h^{2} + k^{2}

⇒ AP = \(\sqrt{h^{2}+k^{2}}\)

Similarly, AQ = \(\sqrt{h^{2}+k^{2}}\)

Again, PQ = \(\sqrt{(h-k)^{2}+(k+h)^{2}}\)

= \(\sqrt{(2k)^{2}}\) = 2k

∴ AP = PQ

⇒ \(\sqrt{h^{2}+k^{2}}\) = 2k

⇒ h^{2} + k^{2} = 4k^{2}

⇒ h^{2} = 3k^{2}

⇒ h = \(\sqrt {3}\).k

∵ Point P(h, k) lies on parabola y^{2} = 4ax.

k^{2} = 4ah = 4a.\(\sqrt {3}\)k

⇒ k = 4a\(\sqrt {3}\), [∵ k ≠ 0]

Hence, length of side PQ = 2k = 2.(4a\(\sqrt {3}\)) = 8a\(\sqrt {3}\).

Question 11.

If a parabola reflector is 20 cm in diameter and 5 cm deep. Find the focus. (NCERT)

Solution:

Taking vertex of parabola reflector at origin and X – axis along the axis of parabola.

Equation of parabola y^{2} = 4ax …. (1)

Given : OS = 5 cm, AB = 20 cm, AS = 10 cm

∴ Co – ordinate of A will be (5, 10).

∴ (10)^{2} = 4a x 5

⇒ 100 = 20a

⇒ a = 5

∴ OS = 5 cm

Co – ordinates of focus S will be (5, 0).

Question 12.

An arch is in the form of a parabola with its vertical axis. The arch is 10 m high and 5 m wide at the base. How wide it is 2 m vertex of the parabola. (NCERT)

Solution:

Let the equation of parabola is :

x^{2} = 4 ay …. (1)

Given : AB = 5 metre

AF = BF = \(\frac { 5 }{ 2 }\) metre

OE = 2 metre

OF = 10 metre

Co – ordinate of A will be (\(\frac { 5 }{ 2 }\), 10)

This point lies on parabola, hence it will be satisfy equation (1),

∴ \(\frac { 5 }{ 2 }\)^{2} = 4a x 10

⇒ \(\frac { 25 }{ 4 }\) = 4 x a x 10

⇒ a = \(\frac { 25 }{ 4 × 4 × 10 }\)

⇒ a = \(\frac { 5 }{ 32 }\)

Put the value of a in Equation (1),

∴ x^{2} = 4 x \(\frac { 5 }{ 32 }\)y

⇒ x^{2} = \(\frac { 5 }{ 8 }\)y

Let EC = k

OE = 2

Co – ordinate of C will be (k, 2) and it will satisfy equation of parabola.

We get k^{2} =\(\frac { 5 }{ 8 }\) x 2

⇒ k^{2} = \(\frac { 5 }{ 4 }\)

⇒ k = \(\frac{\sqrt{5}}{2}\)

DE = 2EC

2 x \(\frac{\sqrt{5}}{2}\) = \(\sqrt {5}\)

= 2.23 metre (approx.)

Question 13.

In each of the following ellipse. Find the co – ordinates of the foci and vertices, the length of major axis and minor axis, the eccentricity and the length of latus rectum of the ellipse. (NCERT)

(A) = \(\frac{x^{2}}{36}+\frac{y^{2}}{16}\) = 1.

Solution:

Comparing with standard form of ellipse, \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1

We get, a^{2} = 36 ⇒ a = 6, b^{2} = 16 ⇒ b = 4

Here a > b.

∴ b^{2} = a^{2}(1 – e^{2})

⇒ 16 = 36(1 – e^{2})

⇒ 1 – e^{2} = \(\frac { 16 }{ 36 }\) = \(\frac { 4 }{ 9 }\)

⇒ e^{2} = 1 – \(\frac { 4}{9 }\) = \(\frac { 5 }{ 9 }\)

∴ Eccentricity e = \(\frac{\sqrt{5}}{3}\)

Foci (± ae, o) = (± 6 × \(\frac{\sqrt{5}}{3}\), o)

= (± 2\(\sqrt {5}\), 0)

Vertices (± a, 0) = (± 6, 0)

Length of major axis = 2a = 2 x 6 = 12.

Length of minor axis = 2b = 2 x 4 = 8.

Length of latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac { 2 × 16 }{ 6 }\) = \(\frac { 16 }{ 3 }\).

(B) \(\frac{x^{2}}{4}+\frac{y^{2}}{25}\) = 1.

Solution:

Comparing with standard form of ellipse, \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1

We get, a^{2} = 36 ⇒ a = 6, b^{2} = 16 ⇒ b = 4

Here a < b.

∴ a^{2} = b^{2}(1 – e^{2})

⇒ 4 = 25(1 – e^{2})

⇒ 1 – e^{2} = \(\frac { 4 }{ 25 }\)

⇒ e^{2} = 1 – \(\frac { 4}{25 }\) = \(\frac { 21 }{ 25 }\)

∴ Eccentricity e = \(\frac{\sqrt{21}}{5}\)

Foci (0, ± b) = (0, ± 5 × \(\frac{\sqrt{21}}{5}\))

= (0, ± \(\sqrt {21}\))

Vertices (0, ± b) = (0, ± 5)

Length of major axis = 2b = 2 x 5 = 12.

Length of minor axis = 2a = 2 x 2 = 8.

Length of latus rectum = \(\frac{2 a^{2}}{b}\) = \(\frac{2 \times 2^{2}}{5}\) = \(\frac { 2 × 4 }{ 5 }\) = \(\frac { 8 }{ 5 }\).

Question 14.

Find the equation of hyperbola whose foci is (± 4, 0) and length of latus rectum is 12. (NCERT)

Solution:

Foci of hyperbola (± 4, 0).

Hence equation of hyperbola will be :

\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1

Foci (±ae, 0) = (± 4, 0)

∴ ae = 4

Length of latus rectum = \(\frac{2 b^{2}}{a}\) = 12

⇒ b^{2} = 6a

We know, b^{2} = a^{2}(e^{2} – 1)

⇒ 6a = a^{2}e^{2} – a^{2}

⇒ 6a = 4^{2} – a^{2}

⇒ 6a = 16 – a^{2}

⇒ 6a = 16 – a^{2}

⇒ a^{2} + 6a – 16 = 0

⇒ a^{2} + 8a – 2a – 16 = 0

⇒ a(a – 2)(a + 8) = 0

⇒ a = 2, a = – 8, (∵ a cannot be negative)

∴ a = 2

b^{2} = 6a = 6 x 2 = 12

⇒ b = \(\sqrt {12}\)

Putting values of a and b in equation (1), then required equation of hyperbola will be :

\(\frac{x^{2}}{2^{2}}-\frac{y^{2}}{(\sqrt{12})^{2}}\) = 1

⇒ \(\frac{x^{2}}{4}+\frac{y^{2}}{12}\) = 1

⇒ 3x^{2} – y^{2} = 12.

Question 15.

Find the axis, foci, directrix, eccentricity and the latus rectum of the ellipse 9x^{2} + 4y^{2} = 36.

Solution:

Given equation of ellipse

9x^{2} + 4y^{2} = 36

⇒ \(\frac{x^{2}}{4}+\frac{y^{2}}{9}\) = 1

Here, b^{2} > a^{2} or b > a

∴ Major axis = 2.3 = 6

Minor axis = 2.2 = 4

Now,

a^{2} = b^{2}(1 – e^{2})

(2)^{2} = (3)^{2}(1 – e^{2})

⇒ 4 = 9(1 – e^{2})

⇒ \(\frac { 4 }{ 9 }\) = 1 – e^{2}

⇒ e^{2} = 1 – \(\frac { 4 }{ 9 }\) = \(\frac { 9 – 4 }{ 9 }\) = \(\frac { 5 }{ 9 }\)

∴ e = \(\frac{\sqrt{5}}{3}\)

Co – ordinate of foci = (0, ± be )

= (0, ± 3. \(\frac{\sqrt{5}}{3}\))

= (0, ± \(\sqrt {5}\)).

Co – ordinate of vertex = (0, ± b )

= (0 ± 3 ).

Length of latus rectum = \(\frac{2 a^{2}}{b}\)

= \(\frac { 2.4 }{ 3 }\) = \(\frac { 8 }{ 3 }\).

Question 16.

(A) Find the equation of ellipse whose vertices are (± 5, 0) and foci (± 4, 0).

Solution:

Given : Vertices are (± 5, 0) and foci are (± 4, 0)

∴ a = 5

and ae = 4

⇒ 5e = 4

⇒ e = \(\frac { 4 }{ 5 }\)

Let the equation of ellipse is :

\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1, where a > b …. (1)

∴ From b^{2} = a^{2}(1 – e^{2}),

b^{2} = 5^{2}(1 – (\(\frac { 4 }{ 5 }\))^{2})

b^{2} = 25(1 – \(\frac { 16 }{ 25 }\))

b^{2} = 25 x \(\frac { 9 }{ 25 }\) = 9

Putting values of a and b in equation (1), the required equation of ellipse will be :

\(\frac{x^{2}}{25}+\frac{y^{2}}{9}\) = 1

⇒ 9x^{2} + 25y^{2} = 225

(B) Find the equation of ellipse whose vertices are (0, ± 13) and foci is (0, ± 5).

Solution:

Let the equation of ellipse is :

\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1, where a < b …. (1)

Vertices of ellipse = (0, ± b) = (0, ± be)

∴ b = 13

Foci = (0, ± 5) = (0, ± be)

∴ be = 15

⇒ 13 x 3 = 5

⇒ e = \(\frac { 5 }{ 13 }\)

Now, a^{2} = b^{2}(1 – e^{2})

⇒ a^{2} = 13^{2}[1 – ( \(\frac { 5 }{ 13 }\))^{2} ]

⇒ a^{2} = 169[1 – \(\frac { 25 }{ 169 }\) ]

⇒ a^{2} = 169\(\frac { 169 – 25 }{ 169 }\)

⇒ a^{2} = 144

⇒ a = 12.

Putting values of a and b in equation (1), then required equation of ellipse will be :

\(\frac{x^{2}}{(12)^{2}}-\frac{y^{2}}{(13)^{2}}\) = 1

⇒ \(\frac{x^{2}}{144}+\frac{y^{2}}{169}\) = 1

Question 17.

Find the equation of ellipse whose centre is at (0, 0), major axis on the Y – axis passing through the points (3, 2) and (1, 6).

Solution:

Let the equation of ellipse is :

\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1, where a < b …. (1)

∵ Equation (1) passes through points(3, 2) and (1, 6)

∴ \(\frac{9}{a^{2}}+\frac{4}{b^{2}}\) = 1 …. (2)

and \(\frac{1}{a^{2}}+\frac{36}{b^{2}}\) = 1 …. (3)

Multiply equation (2) by 9, we get

Putting value of a^{2} in equation (2), we get

\(\frac { 9 }{ 10 }\) + \(\frac{4}{b^{2}}\) = 1

⇒ \(\frac{4}{b^{2}}\) 1 – \(\frac { 9 }{ 10 }\)

⇒ \(\frac{4}{b^{2}}\) = \(\frac { 1}{ 10 }\)

⇒ b^{2} = 40

Putting values of a^{2} and b^{2} in equation (1), then required equation of ellipse will be :

\(\frac{x^{2}}{10}+\frac{y^{2}}{40}\) = 1

Question 18.

Find the equation of ellipse whose major axis on the X – axis which passes through the points (4, 3) and (6, 2).

Solution:

Let the equation of ellipse is :

\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 …. (1)

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

∴\(\frac{36}{a^{2}}+\frac{4}{b^{2}}\) = 1 …. (2)

and \(\frac{16}{a^{2}}+\frac{9}{b^{2}}\) = 1 …. (3)

Subtracting equation (3) from equation (2), we get

\(\frac{20}{a^{2}}-\frac{5}{b^{2}}\) = 1

⇒ \(\frac{4}{a^{2}}-\frac{1}{b^{2}}\) = 1

⇒ \(\frac{4}{a^{2}} = \frac{1}{b^{2}}\)

⇒ a^{2} = 4b^{2}

Putting value of a^{2} in equation (2), we get

\(\frac{36}{4b^{2}}+\frac{4}{b^{2}}\) = 1

⇒ \(\frac{9}{b^{2}}+\frac{4}{b^{2}}\) = 1

⇒ 9 + 4 = b^{2}

⇒ b^{2} = 13

Putting values of a^{2} and b^{2} in equation (1), hence

Required equation of ellipse \(\frac{x^{2}}{52}+\frac{y^{2}}{13}\) = 1

Question 19.

An arch is the form of a semi ellipse. It is 8 m wide and 2 m high of the centre. Find the height of the arch at a point 1-5 m from one end. (NCERT)

Solution:

Let the equation of ellipse is :

\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 …. (1)

Here 2a = 8 ⇒ a = 4, b = 2

Putting the values of a and b, we get

\(\frac{x^{2}}{4^{2}}-\frac{y^{2}}{2^{2}}\) = 1 …. (1)

\(\frac{x^{2}}{16}+\frac{y^{2}}{4}\) = 1

Given :

AP = 1.5m, OA = \(\frac { 8 }{ 2 }\) = 4m,

OP = OA – AP = 4 – 1.5 = 2.5m.

Let PQ = k

∴ Co – ordinate of Q will be (2.5, k), which will satisfy ellipse’s equation.

Hence,

\(\frac{(2.5)^{2}}{16}+\frac{k^{2}}{4}\) = 1

⇒ \(\frac { 6.25}{ 16 }\) + \(\frac{k^{2}}{4}\) = 1

⇒ \(\frac{k^{2}}{4}\) = 1 – \(\frac { 6.25 }{ 16 }\)

⇒ \(\frac{k^{2}}{4}\) = \(\frac { 16 – 6.25 }{ 16 }\)

⇒ k^{2} = \(\frac { 9.75 }{ 4 }\)

⇒ k^{2} = 2.437

⇒ k = 1.56metre (approx).

Question 20.

A rod of length 12 cm moves with its ends always touching the co – ordinate 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. (NCERT)

Solution:

Let AB be the rod of length 12 cm which make an angle θ with X – axis.

∴ ∠BAO = θ

AB = 12 cm

AP = 3 cm, then PB = 9 cm

In ∆PNA,

sin θ = \(\frac {PN }{ PA }\) = \(\frac { y }{ 3 }\)

In ∆PMB,

cos θ = \(\frac { PM }{ PB }\) = \(\frac { x }{ 9 }\)

sin^{2}θ + cos^{2}θ = \(\frac { y }{ 3 }\)^{2} + \(\frac { x }{ 9 }\)^{2}

⇒ 1 = \(\frac{y^{2}}{9}+\frac{x^{2}}{81}\)

Hence required equation is :

\(\frac{x^{2}}{81}+\frac{y^{2}}{9}\) = 1

Question 21.

Find the eccentricity, co – ordinate of foci, equation of directrix and length of Iatus rectum of ellipse 4x^{2} + y^{2} – 8x + 2y + 1 = 0.

Solution:

4x^{2} + y^{2} – 8x + 2y + 1 = 0

⇒ 4x^{2} – 8x + y^{2} + 2y +1 = 0

⇒ 4x^{2} – 8x + (y + 1)^{2} = 0

⇒ 4(x^{2} – 2x) + (y + 1)^{2} = 0

⇒ 4(x^{2} – 2x + 1) + (y + 1)^{2} = 4

⇒ 4(x^{2} – 2x + 1) + (y + 1)^{2} = 4

⇒ \(\frac{(x-1)^{2}}{1}+\frac{(y+1)^{2}}{4}\) or \(\frac{X^{2}}{1}+\frac{Y^{2}}{4}\) = 1

Here, b > a

a^{2} = b^{2}(1 – e^{2})

⇒ 1 = (1 – e^{2})

⇒ \(\frac { 1 }{ 4}\) = 1 – e^{2}

⇒ e^{2} = 1 – \(\frac { 1 }{ 4}\) = \(\frac { 3 }{ 4}\)

⇒ Eccentricity e = \(\frac{\sqrt{3}}{2}\)

Co – ordinate of foci (0, ± be)

= (0, ±2. \(\frac{\sqrt{3}}{2}\)

= (0, ± \(\sqrt {3}\) )

Here X = 0, Y = ± \(\sqrt {3}\)

∴ x – 1 = 0, y + 1 = ± \(\sqrt {3}\)

⇒ x = 1, y = – 1 ± \(\sqrt {3}\)

foci = (1 ± \(\sqrt {3}\) – 1)

Equation of directrix Y = ± \(\frac { b }{ e}\)

⇒ Y = ± \(\frac{2}{\sqrt{3}}\).2

⇒ Y = ± \(\frac{4}{\sqrt{3}}\)

Y + 1 = \(\frac{4}{\sqrt{3}}\), (∵ Y = y+1)

⇒ y = ± \(\frac{4}{\sqrt{3}}\) – 1

Length of latus rectum = \(\frac{2 a^{2}}{b}\)

= 2. \(\frac { 1 }{ 2 }\) = 1.

Question 22.

Find the vertices, co-ordinate of foci, eccentricity and length of latus rectum of hyperbola :

(A) 9y^{2} – 4x^{2} = 36.

Solution:

Given : 9y^{2} – 4x^{2} = 36

⇒ \(\frac{9 y^{2}}{36}-\frac{4 x^{2}}{36}\) = 1

⇒ \(\frac{y^{2}}{4}-\frac{x^{2}}{9}\) = 1

Comparing the above equation with the standard form of hyperbola

\(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}\) = 1 …. (1)

b^{2} = 4 ⇒ b = 2, a^{2} = 9 ⇒ a = 3

Let e is the eccentricity of hyperbola.

Then, a^{2} = b^{2}(e^{2} – 1)

⇒ 9 = (e^{2} – 1)

⇒ \(\frac { 9 }{ 4 }\) = e^{2} – 1

⇒ e^{2} = \(\frac { 9 }{ 4 }\) + 1 = \(\frac { 13 }{ 4 }\)

∴ Eccentricity e = \(\frac{\sqrt{3}}{2}\)

Vertices = (0, ± b) = (0, ± 2)

Foci = (0, ± be) = (0, ± 2 x \(\frac{\sqrt{3}}{2}\)) = (0, ±\(\sqrt {3}\))

Length of latus rectum = \(\frac{2 a^{2}}{b}\) = \(\frac { 2 x 9 }{ 2 }\) = 9

(B) 16x^{2} – 9y^{2} = 576.

Solution:

Given : 16x^{2} – 9y^{2} = 576

⇒ \(\frac{16 x^{2}}{576}-\frac{9 y^{2}}{576}\) = 1

⇒ \(\frac{x^{2}}{36}-\frac{y^{2}}{64}\) = 1

Comparing the above equation with the standard form of hyperbola

\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 …. (1)

Here, a^{2} = 36 ⇒ a = 6, b^{2} = 64 ⇒ b = 3

We Know that, b^{2} = a^{2}(e^{2} – 1)

64 = 36(e^{2} – 1)

⇒ \(\frac { 64 }{ 36 }\) = e^{2} – 1

⇒ \(\frac { 16 }{ 9 }\) = e^{2} – 1

⇒ e^{2} = \(\frac { 16 }{ 9 }\) + 1 = \(\frac { 25 }{ 9 }\)

∴ Eccentricity e = \(\frac{5}{3}\)

Vertices = (± a, 0) = (± 6, 0)

Foci = (± ae, 0) = (± 6 x \(\frac { 5 }{ 3 }\)) = (± 10, 0)

Length of latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac { 2 x 64 }{ 6 }\) = \(\frac { 64 }{ 3 }\).

(c) 5y^{2} – 9x^{2} = 36.

Solution:

Given : 5y^{2} – 9x^{2} = 36

⇒ \(\frac{5 y^{2}}{36}-\frac{9 x^{2}}{36}\) = 1

\(\frac{y^{2}}{\frac{36}{5}}-\frac{x^{2}}{4}\) = 1

Comparing the above equation with the standard form of hyperbola

\(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}\) = 1 …. (1)

Here b^{2} = \(\frac { 36 }{ 5 }\) ⇒ b = \(\frac{\sqrt{6}}{5}\), a^{2} = 4 ⇒ a = 2

We know that, a^{2} = b^{2}(e^{2} – 1)

⇒ 4 = \(\frac{\sqrt{36}}{5}\)(e^{2} – 1)

⇒ e^{2} – 1 = \(\frac { 20 }{ 36 }\) = \(\frac { 5 }{ 9 }\)

⇒ e^{2} = 1 + \(\frac { 5 }{ 9 }\) = \(\frac { 14 }{ 9 }\)

Question 23.

Find the equation of hyperbola whose foci are (0, ± \(\sqrt {10}\)) and which passes through point (2,3).

Solution:

Foci of hyperbola are (0, ±\(\sqrt {10}\)).

∴Form of hyperbola is :

\(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}\) = 1 …. (1)

Foci (0, ± be) = (0, ± \(\sqrt {10}\))

be = \(\sqrt {10}\)

Equation (1) passes through point (2, 3).

∴ \(\frac{9}{b^{2}}-\frac{4}{a^{2}}\)

⇒ 9a^{2} – 4b^{2} = a^{2}b^{2}

We know that, a^{2} = b^{2} (e^{2} – 1)

⇒ a^{2} = b^{2}e^{2} – b^{2}

a^{2} = ( \(\sqrt {10}\))^{2} – b^{2}

⇒ a^{2} = 10 – b^{2}

⇒ b^{2} = 10 – a^{2}

Putting value of b^{2} in equation (2),

9a^{2} – 4(10 – a^{2}) = a^{2} (10 – a^{2})

⇒ 9a^{2} – 40 + 4a^{2} = 10a^{2} – a^{4}

⇒ 13a^{2} – 40 = 10a^{2} – a^{4}

⇒ a^{4} + 13a^{2} – 10a^{2} – 40 = 0

⇒ a^{4} + 3a^{2} – 40 = 0

⇒ a^{4} + 8a^{2} – 5a^{2} – 40 = 0

⇒ a^{2}(a^{2} + 8) – 5(a^{2} + 8) = 0

⇒ (a^{2} – 5)(a^{2} + 8) = 0

a^{2} = 5, a^{2} = – 8

∵ The value of a cannot be negative.

∴ a^{2} = 5

b^{2} = 10 – a^{2}

⇒ b^{2} = 10 – 5

⇒ b^{2} = 5

Putting values of a^{2} and b^{2} in equation (1), then required equation of hyperbola will be :

\(\frac{y^{2}}{5}-\frac{x^{2}}{5}\) = 1

⇒ y^{2} – x^{2} = 5.

Question 24.

Find the equation of hyperbola in which the distance between foci is 8 and distance between directrix is 6.

Solution:

Let the equation of hyperbola is :

\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) … (1)

and Eccentricity of hyperbola is e and foci are (ae, 0) and (- ae, 0) and latus rectum

are x = \(\frac { a }{ e }\) and x = – \(\frac { a }{ e }\)

Distance between foci = 2ae

Distance between latus rectum = \(\frac { 2a }{ e }\)

According to question, 2ae = 8

and \(\frac { 2a }{ e }\) = 6

Multiplying equation (2) and (3),

4a^{2} = 48 ⇒ a^{2} = 12

⇒ a = 2\(\sqrt {3}\)

Putting value of a in equation (2),

2.2\(\sqrt {3}\)e = 8 ⇒ e = \(\frac{2}{\sqrt{3}}\)

b^{2} = a^{2}(e^{2} – 1)

= 12( \(\frac { 4 }{ 3 }\) – 1) = 4

Putting values of a^{2} and b^{2} in equation (1), the required equation of hyperbola is :

\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\)

⇒ x^{2} – 3y^{2} = 12.

Question 25.

Find the centre, eccentricity, foci and length of latus rectum of hyperbola 9x^{2} – 16y^{2} + 18x + 32y – 151 = 0.

Solution:

Equation of hyperbola is :

9x^{2} – 16y^{2} + 18x + 32y – 151 = 0

⇒ 9x^{2} + 18x – 16y^{2} + 32y = 151

⇒ 9(x^{2} + 2x) – 16(y^{2} – 2y) = 151

⇒ 9(x + 1)^{2} – 16(y – 1)^{2} = 151 – 16 + 9

⇒ 9(x + 1)^{2} – 16(y – 1)^{2} = 144

⇒ \(\frac{9(x+1)^{2}}{144}-\frac{16(y-1)^{2}}{144}\)

⇒ \(\frac{(x+1)^{2}}{16}-\frac{(y-1)^{2}}{9}\) = 1

Let x + 1 = X and y – 1 = Y, then equation of hyperbola is

\(\frac{X^{2}}{16} – \frac{Y^{2}}{9}\) = 1

∴ a^{2} = 16 ⇒ a = 4 and b^{2} = 9 ⇒ b = 3.

∴ Centre is (- 1, 1).

For eccentricity = e

b^{2} = a^{2}(e^{2} – 1)

⇒ 9 = 16(e^{2} – 1)

⇒ \(\frac { 9 }{ 16 }\) = e^{2} – 1

⇒ e^{2} = 1 + \(\frac { 9 }{ 16 }\) = \(\frac { 25 }{ 16 }\)

⇒ e = \(\frac { 5 }{ 4 }\)

For foci, X = ± ae, Y = 0

⇒ x +1 = ± 4 x \(\frac { 5 }{ 4 }\), y – 1 = 0

⇒ x + 1 = ± 5, y = 1

⇒ x = 4, – 6, y = 1

∴ Foci are (4, 1) and (6, 1).

Equation of directrix is X = ± \(\frac { a }{ e }\)

⇒ x +1 = ± \(\frac { 4 }{ 5/4 }\)

⇒ x = ± \(\frac { 16 }{ 5 }\) – 1

⇒ x = ± \(\frac { 11 }{ 5 }\) and x = – \(\frac { 21 }{ 5 }\)

⇒ 5x = 11 and 5x + 21 = 0.

Question 26.

If e and ex are the eccentricity of hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) and \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}\) = 1, then prove that: \(\frac{1}{e^{2}}+\frac{1}{e_{1}^{2}}\) = 1.

Solution:

Equation of hyperbola is

\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1

and

\(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}\) = 1

For eccentricity e of equation (1),

b^{2} = a^{2}(e^{2} – 1)

⇒ \(\frac{b^{2}}{a^{2}}\) = (e^{2} – 1)

⇒ e ^{2} = 1 + \(\frac{b^{2}}{a^{2}}\) = \(\frac{a^{2}+b^{2}}{b^{2}}\)

⇒ \(\frac{1}{e_{1}^{2}}\) = \(\frac{a^{2}}{a^{2}+b^{2}}\)

Again for eccentricity e of equation (2),

Question 27.

On a level plain the crack of the rifle and the thud of the ball striking the target are heard at the same instant, prove that the locus of the hearer is a hyperbola.

Solution:

Let P be the situation of hearer and T be the situation of the rifle and S is target. Let the velocity of the ball be v_{1} and the velocity of sound be v_{2}.

Then,

Time to reach the ball from T to S = \(\frac{T S}{v_{1}}\)

Time to reach the sound from S to P = \(\frac{S P}{v_{2}}\)

and Time to reach the sound from T to P = \(\frac{T P}{v_{2}}\)

∴ The crack of the rifle and the thud of the ball are heard at the same instant.

∴ \(\frac{T S}{v_{1}}\) + \(\frac{S P}{v_{2}}\) = \(\frac{T P}{v_{2}}\)

⇒ \(\frac{T P}{v_{2}}\) – \(\frac{S P}{v_{2}}\) = \(\frac{T S}{v_{1}}\)

⇒ TP – SP = \(\frac{v_{2}}{v_{1}}\)

⇒ PT – PS = A constant (∵ v_{2}, v_{2}, TS are constant)

Hence locus of point P is hyperbola whose foci is T and S.

#### MP Board Class 11th Maths Important Questions