Find the equation of tangents to the ellipse x 2 50 + y 2 32 = 1 , which passes through a point ( 15 , - 4 ) .

Given the equation of ellipse x 2 50 + y 2 32 = 1 . The equation of tangent passing through 15 , - 4 . From ellipse, we have a 2 = 50 , b 2 = 32 . We know the equation of tangent to the ellipse x 2 a 2 + y 2 b 2 = 1 is y = m x ± a 2 m 2 + b 2 . . . 1 So, y - m x = ± a 2 m 2 + b 2 ( passing through 15 , - 4 ) - 4 - m × 15 = ± 50 m 2 + 32 Squaring both sides, 16 + 225 m 2 + 120 m = 50 m 2 + 32 175 m 2 + 120 m - 16 = 0 = m + 4 5 m - 4 35 So, m = - 4 5 , 4 35 The equation of the tangent to the ellipse x 2 a 2 + y 2 b 2 = 1 is y = m x ± a 2 m 2 + b 2 Case 1 for m = - 4 5 y = - 4 5 x + 50 × 16 25 + 32 ⇒ 5 y + 4 x = 1600 ⇒ 4 x + 5 y = 40 Case 2 for m = 4 35 y = 4 35 x + 4 35 2 × 50 + 32 ⇒ 35 y = 4 x + 4000 ⇒ 4 x - 35 y = 200

If the normal at a point P on the ellipse x 2 a 2 + y 2 b 2 = 1 of semi axes a , b centre C cuts the major & minor axes at G & g , show that a 2 · C G 2 + b 2 · C g 2 = a 2 - b 2 2 . Also prove that C G = e 2 C N , where P N is the ordinate of P . ( N is foot of perpendicular from P on its major axis )

Let equation of ellipse is x 2 a 2 + y 2 b 2 = 1 . Let point is P ( a cos θ , b sin θ ) . Equation of normal at P is a x sec θ - b y cosec θ = a 2 - b 2 ⇒ x a 2 - b 2 cos θ a + y - a 2 - b 2 sin θ b = 1 It meet major and minor axis at G a 2 - b 2 a cos θ , 0 and g 0 , - a 2 - b 2 b sin θ respectively. ∴ C G 2 = a 2 - b 2 a 2 cos 2 θ and C g 2 = a 2 - b 2 b 2 sin 2 θ ∴ a 2 C G 2 + b 2 C g 2 = a 2 - b 2 2 PN is ordinate, ∴ coordinate of N ( acos θ , 0 ) . e 2 C N = a 2 - b 2 a 2 a cos θ = a 2 - b 2 a cos θ = C G

A tangent to the ellipse x 2 + 4 y 2 = 4 meets the ellipse x 2 + 2 y 2 = 6 at P & Q . Prove that the tangents at P & Q of the ellipse x 2 + 2 y 2 = 6 are at right angles.

Given, x 2 + 4 y 2 = 4 ⇒ x 2 4 + y 2 1 = 1 . . . 1 x 2 + 2 y 2 = 6 ⇒ x 2 6 + y 2 3 = 1 . . . 2 Let T be ( h , k ) As, P Q is the chord of contact of T to the ellipse which is given in Eq. (2) Hence, h x 6 + k y 3 = 1 ⇒ y = - h 2 k x + 3 k . . . 3 As, it touches the ellipse (1), hence, we get, ⇒ 9 k 2 = 4 - h 2 k + 1 ⇒ h 2 + k 2 = 9 = 6 + 3 Now replacing, h → x & k → y x 2 + y 2 = 9 Therefore, T ( h , k ) lies on the director circle of Eq. (2) so that the tangents at P & Q are at right angles.

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The tangent at the point $α$ on a standard ellipse meets the auxiliary circle in two points which subtends a right angle at the centre. Show that the eccentricity of the ellipse is ${(1 + \sin^{2} α)}^{- \frac{1}{2}}$ .

Important Questions on Ellipse

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 19 - Ellipse>EXERCISE-4>Q 7

Find the equation of tangents to the ellipse

\frac{x^{2}}{50} + \frac{y^{2}}{32} = 1

, which passes through a point

(15, - 4)

EASY

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 19 - Ellipse>EXERCISE-4>Q 8

A B C

is an isosceles triangle with its base

B C

twice its altitude. A point

P

moves within the triangle such that the square of its distance from

B C

is half the area of rectangle contained by its distances from the two sides. Show that the locus of

P

is an ellipse with eccentricity

\sqrt{\frac{2}{3}}

passing through

B & C

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 19 - Ellipse>EXERCISE-4>Q 9

If the normal at a point

P

on the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

of semi axes

a, b

centre

C

cuts the major & minor axes at

G & g

, show that

a^{2} \cdot {(C G)}^{2} + b^{2} \cdot {(C g)}^{2} = {(a^{2} - b^{2})}^{2}

. Also prove that

C G = e^{2} C N

, where

P N

is the ordinate of

P

(N

is foot of perpendicular from

P

on its major axis

)

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 19 - Ellipse>EXERCISE-4>Q 10

A ray emanating from the point

(- 4, 0)

is incident on the ellipse

9 x^{2} + 25 y^{2} = 225

at the point

P

with abscissa

3 .

Find the equation of the reflected ray after first reflection.

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JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 19 - Ellipse>EXERCISE-4>Q 11

Find the locus of the point, the chord of contact of the tangents drawn from which to the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

touches the circle

x^{2} + y^{2} = c^{2}

, where

c < b < a

EASY

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 19 - Ellipse>EXERCISE-4>Q 12

A tangent to the ellipse

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

meets the ellipse

x^{2} + 2 y^{2} = 6

P & Q

. Prove that the tangents at

P & Q

of the ellipse

x^{2} + 2 y^{2} = 6

are at right angles.

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 19 - Ellipse>EXERCISE-4>Q 13

The tangent at any point

P

of a circle

x^{2} + y^{2} = a^{2}

meets the tangent at a fixed point

A (a, 0)

T

and

T

is joined to

B

, the other end of the diameter through

A

, Prove that the locus of intersection of

A P

and

B T

is an ellipse whose eccentricity is

\frac{1}{\sqrt{2}}

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 19 - Ellipse>EXERCISE-4>Q 14

The tangent at

P (4 cosθ, \frac{16}{\sqrt{11}} sinθ)

to the ellipse

16 x^{2} + 11 y^{2} = 256

is also a tangent to the circle

x^{2} + y^{2} - 2 x - 15 = 0 .

Find

θ .

Find also the equation to the common tangent.

The tangent at the point α on a standard ellipse meets the auxiliary circle in two points which subtends a right angle at the centre. Show that the eccentricity of the ellipse is 1+sin2α-12.

Important Questions on Ellipse

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The tangent at the point $α$ on a standard ellipse meets the auxiliary circle in two points which subtends a right angle at the centre. Show that the eccentricity of the ellipse is ${(1 + \sin^{2} α)}^{- \frac{1}{2}}$ .