Show that the angle between the tangents to the ellipse x 2 a 2 + y 2 b 2 = 1 a > b > 0 and the circle x 2 + y 2 = a b at their points of intersection is t a n - 1 a - b a b .

The equation of the ellipse is: x 2 a 2 + y 2 b 2 = 1 … i and equation of the circle is x 2 + y 2 = a b . . . ii Let the point of the intersection in the first quadrant is P a b cos θ , a b sin θ , this point will satisfy the ellipse ∴ a b cos θ 2 a 2 + a b sin θ 2 b 2 = 1 ⇒ a b cos 2 θ a 2 + a b sin 2 θ b 2 = 1 ⇒ b 1 - sin 2 θ a + a sin 2 θ b = 1 ⇒ sin θ = b a + b . . . iii for point of intersection in first quadrant. Equation of tangent at P on the ellipse is x a b cos θ a 2 + y a b sin θ b 2 = 1 Let its slope, m 1 = - b 2 a 2 cos θ sin θ . . . i v Similarly, equation of tangent at P on the circle is a b cos θ x + a b sin θ y = a b ⇒ cos θ x + sin θ y = a b Its slope, m 2 = - cos θ sin θ . . . v Hence, if θ be the angle between these two tangents at P , then t a n θ = m 1 - m 2 1 + m 1 m 2 ; Putting the values of m 1 and m 2 , ∴ tan θ = - b 2 a 2 cos θ sin θ + cos θ sin θ 1 + b 2 a 2 cos θ sin θ cos θ sin θ ⇒ tan θ = 1 - b 2 a 2 cot θ 1 + b 2 a 2 cot 2 θ . . . vi ; From equation ( iii ) , we have sin θ = b a + b , so cot θ = a b ; Now, putting the value of cot θ in equation v i , we get tan θ = 1 - b 2 a 2 a b 1 + b 2 a 2 a b 2 ∴ θ = tan - 1 a - b a b . Hence proved.

Prove that the common tangent of the ellipses x 2 a 2 + y 2 b 2 = 2 x c and x 2 b 2 + y 2 a 2 + 2 x c = 0 subtends a right angle at the origin.

The equation to the ellipses are given as x 2 a 2 + y 2 b 2 = 2 x c . . . i and x 2 b 2 + y 2 a 2 + 2 x c = 0 . . . ii Equations i and ii can be rewritten as x a - a c 2 + y 2 b 2 - a 2 c 2 = 0 and x b + b c 2 + y 2 a 2 - b 2 c 2 = 0 . The line y = a b c is a common tangent (Using the concept a 2 + b 2 = 0 , when a = 0 & b = 0 ) Since, it cuts the two ellipses in coincident points whose co-ordinates are P a 2 c , a b c and Q - b 2 c , a b c , respectively, and the co-ordinates of C are 0, 0 . Let slope of P C is = m 1 = a b c a 2 c = b a And slope of Q C is = m 2 = a b c - b 2 c = - a b ⇒ m 1 . m 2 = - 1 Hence, common tangent will subtend right angle at origin.

HARD

JEE Advanced

IMPORTANT

Earn 100

Two tangents to an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b)$ intersect at right angles. Prove that the sum of squares of the chords which the auxiliary circles intercepts on them is constant and equal to square of the length of the line joining foci of the ellipse.

Important Questions on The Ellipse

HARD

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 10 - The Ellipse>EXAMPLES XXXIII>Q 13

P

is a point on an ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b > 0)

whose ordinate is

y^{'}

, prove that the angle between the tangent at

P

and the corresponding focal distance of

P

t a n^{- 1} \frac{b^{2}}{a e y^{'}}

HARD

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 10 - The Ellipse>EXAMPLES XXXIII>Q 14

Show that the angle between the tangents to the ellipse

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

(a > b > 0)

and the circle

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

at their points of intersection is

t a n^{- 1} (\frac{a - b}{\sqrt{a b}})

HARD

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 10 - The Ellipse>EXAMPLES XXXIII>Q 15

A circle of radius

r

is concentric with an ellipse

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

; prove that the common tangent is inclined to the major axis at an angle

t a n^{- 1} (\pm \sqrt{\frac{r^{2} - b^{2}}{a^{2} - r^{2}}})

and find its length.

HARD

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 10 - The Ellipse>EXAMPLES XXXIII>Q 16

Prove that the common tangent of the ellipses

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

and

\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} + \frac{2 x}{c} = 0

subtends a right angle at the origin.

HARD

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 10 - The Ellipse>EXAMPLES XXXIII>Q 17

At any point $P$ on the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b))$ . A Tangent and Normal drawn to it. The tangent and normal intersects the $x$ -axis at $T$ and $G$ respectively, the normal intersects the $y$ -axis at $g,$ Prove that $PG . Pg = SP . S^{'} P$ and $CG . CT = {CS}^{2},$ Where $C$ is the centre and $S$ and $S'$ are foci of the ellipse.

HARD

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 10 - The Ellipse>EXAMPLES XXXIII>Q 18

The tangent to the ellipse at

P

meets the axis in

T

and

t,

and

C Y

is the perpendicular on it from the centre; prove that

T t . P Y = a^{2} - b^{2} .

HARD

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 10 - The Ellipse>EXAMPLES XXXIII>Q 18

The tangent at $P$ on the ellipse meets the axis in $T$ and $t$ , and $C Y$ is the perpendicular on it from the center; prove that $(i) T t . P Y = a^{2} - b^{2}$ , and $(ii)$ the least value of $T t$ is $a + b$ .

HARD

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 10 - The Ellipse>EXAMPLES XXXIII>Q 19

Prove that the perpendicular from the focus upon any tangent on the ellipse and the line joining the center to the point of contact, meet on the corresponding directrix.

Two tangents to an ellipse x2a2+y2b2=1(a>b) intersect at right angles. Prove that the sum of squares of the chords which the auxiliary circles intercepts on them is constant and equal to square of the length of the line joining foci of the ellipse.

Important Questions on The Ellipse

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Two tangents to an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b)$ intersect at right angles. Prove that the sum of squares of the chords which the auxiliary circles intercepts on them is constant and equal to square of the length of the line joining foci of the ellipse.