If the normal at P θ and Q π 2 + θ to the ellipses x 2 a 2 + y 2 b 2 = 1 : a > b meet the major axis at G and g , respectively, then the value of P G 2 + Q g 2 is a 2 e 2 - 1 e 2 - 2 .

We have, an ellipse x 2 a 2 + y 2 b 2 = 1 , a > b , hence the major axis of the ellipse is x -axis i.e. y = 0 . Normal at P θ is a x cos θ - b y sin θ = a 2 - b 2 . . . i Normal at Q π 2 + θ is a x cos π 2 + θ - b y sin π 2 + θ = a 2 - b 2 ⇒ - a x sin θ - b y cos θ = a 2 - b 2 . . . i i Equations i and i i meet major axis at G a 2 - b 2 cos θ a , 0 and g - a 2 - b 2 sin θ a , 0 . Here, P ≡ a cos θ , b sin θ and Q ≡ a cos π 2 + θ , b sin π 2 + θ ⇒ Q ≡ - a sin θ , b cos θ . The distance between the points x 1 , y 1 and x 2 , y 2 is x 1 - x 2 2 + y 1 - y 2 2 . Now, P G 2 + Q g 2 = a 2 - b 2 cos θ a - a cos θ 2 + 0 - b sin θ 2 + a 2 - b 2 sin θ a - a sin θ 2 + 0 - b cos θ 2 = a 2 cos θ - b 2 cos θ - a 2 cos θ a 2 + b 2 sin 2 θ + a 2 sin θ - b 2 sin θ - a 2 sin θ a 2 + b 2 cos 2 θ = b 4 cos 2 θ a 2 + b 2 sin 2 θ + b 4 sin 2 θ a 2 + b 2 cos 2 θ = b 4 cos 2 θ + sin 2 θ a 2 + b 2 cos 2 θ + sin 2 θ = b 4 a 2 + a 2 . . . i i i We know that, the eccentricity of the ellipse x 2 a 2 + y 2 b 2 = 1 , a > b is e = 1 - b 2 a 2 . ⇒ e 2 = 1 - b 2 a 2 ⇒ b 2 a 2 = 1 - e 2 ⇒ b 2 = a 2 1 - e 2 . On putting the value of b 2 in the equation i i i , we get b 4 a 2 + a 2 = a 4 1 - e 2 2 a 2 + a 2 = a 2 1 + e 4 - 2 e 2 + a 2 = a 2 e 4 - e 2 + 2 = a 2 e 4 - 2 e 2 - e 2 + 2 = a 2 e 2 e 2 - 2 - e 2 - 2 = a 2 e 2 - 2 e 2 - 1 . Hence, the value of P G 2 + Q g 2 = a 2 e 2 - 1 e 2 - 2 .

An ellipse is rotated through a right angle in its own plane about its center, which is fixed; prove that the locus of the point of intersection of a tangent to the ellipse x 2 a 2 + y 2 b 2 = 1 in its original position, with the tangent at the same point of the curve in its new position is x 2 + y 2 x 2 + y 2 - a 2 - b 2 = 2 a 2 - b 2 x y .

We know that the equation that is always a tangent on the ellipse x 2 a 2 + y 2 b 2 = 1 is given as, x cos α + y sin α = a 2 cos 2 α + b 2 sin 2 α … 1 The equation of the tangent to the ellipse when it has been rotated through 90 ° will be obtained by substituting α + 90 ° as α on the left-hand side of the original tangent represented by equation ( 1 ) . Hence, the equation of the tangent in the second position will be given as, x cos 90 ° + α + y sin 90 ° + α = a 2 cos 2 α + b 2 sin 2 α - x sin α + y cos α = a 2 cos 2 α + b 2 sin 2 α … 2 Subtracting equation 2 from 1 , we get, y + x sin α + x - y cos α = 0 ∴ sin α y - x = cos α y + x = 1 y - x 2 + y + x 2 = 1 2 y 2 + x 2 Substituting the values of sin α and cos α in equation 1 , we get, x y + x + y y - x = a 2 y + x 2 + b 2 y - x 2 x 2 + y 2 = a 2 + b 2 x 2 + y 2 + 2 x y a 2 - b 2 x 2 + y 2 2 = a 2 + b 2 x 2 + y 2 + 2 x y a 2 - b 2 x 2 + y 2 x 2 + y 2 - a 2 - b 2 = 2 x y a 2 - b 2

Prove that the area of the triangle formed by lines joining the points on the ellipse x 2 a 2 + y 2 b 2 = 1 whose eccentric angles are α , β , γ is 1 2 a b sin β - γ + sin α - β + sin γ - α .

We know that, the co-ordinates of a point on the ellipse x 2 a 2 + y 2 b 2 = 1 with eccentric angle θ is a cos θ , b sin θ . Thus, the points on the ellipse x 2 a 2 + y 2 b 2 = 1 whose eccentric angles are α , β , γ are a cos α , b sin α , a cos β , b sin β and a cos γ , b sin γ . The area of the triangle with vertices x 1 , y 1 , x 2 , y 2 and x 3 , y 3 is the absolute value of 1 2 x 1 y 1 1 x 2 y 2 1 x 3 y 3 1 . Thus, the area of the required triangle is ∆ = 1 2 a cos α b sin α 1 a cos β b sin β 1 a cos γ b sin γ 1 . Taking a common from the first column and b from the second column. ⇒ ∆ = a b 2 cos α sin α 1 cos β sin β 1 cos γ sin γ 1 . Applying R 2 → R 2 - R 1 , R 3 → R 3 - R 1 ⇒ ∆ = 1 2 a b cos α sin α 1 cos β - cos α sin β - sin α 0 cos γ - cos α sin γ - sin α 0 ⇒ ∆ = 1 2 a b cos β - cos α sin γ - sin α - sin β - sin α cos γ - cos α ⇒ ∆ = 1 2 a b cos β sin γ - cos α sin γ - cos β sin α + cos α sin α - sin β cos γ + sin α cos γ + sin β cos α - sin α cos α ⇒ ∆ = 1 2 a b cos β sin γ - cos α sin γ - cos β sin α - sin β cos γ + sin α cos γ + sin β cos α ⇒ ∆ = 1 2 a b cos β sin γ - sin β cos γ + sin α cos γ - cos α sin γ + sin β cos α - cos β sin α . Using sin A cos B - cos A sin B = sin A - B , we get ∆ = 1 2 a b sin γ - β + sin α - γ + sin β - α . Also, we know that sin - θ = - sin θ . ⇒ ∆ = 1 2 a b sin β - γ + sin α - β + sin γ - α .

Prove that the directrices of the two parabolas that can be drawn to have their foci at any given point P of the ellipse x 2 a 2 + y 2 b 2 = 1 and to pass through its foci meet at an angle which is equal to twice the eccentric angle of P .

Let the ellipse be x 2 a 2 + y 2 b 2 = 1 having foci as S - a e , 0 and S ' a e , 0 . Take any point P ( a cos ϕ , b sin ϕ ) on it. Then, P S = a + a e cos ⁡ ϕ , and P S ' = a - a e cos ϕ Let the directrix of the parabola be x cos ⁡ α + y sin ⁡ α = p ; Then the point ( a cos ⁡ ϕ , b sin ⁡ ϕ ) is focus of the parabola and by hypothesis, it passes through S and S ' . Therefore, by the definition of the parabola, perpendicular from S on the directrix must be equal to S P . Or p + a e cos ⁡ α = a + a e cos ⁡ ϕ . … 1 Again the perpendicular from S ' on directrix = S ' P So, p - a e cos ⁡ α = a - a e cos ⁡ ϕ . . . 2 Subtract equation ( 2 ) from equation ( 1 ) , we have: cos ⁡ α = cos ⁡ ϕ Hence, α = ϕ or - ϕ ; Therefore, the angle between the directrices is, ϕ - - ϕ = 2 ϕ , which is double of the eccentric angle of P .

Prove that the equation to the circle, having double contact with the ellipse x 2 a 2 + y 2 b 2 = 1 a > b at the ends of a latus rectum, is x 2 + y 2 - 2 a e 3 x = a 2 1 - e 2 - e 4 .

Observe the figure below: Let the equation of the ellipse is: S ≡ x 2 a 2 + y 2 b 2 = 1 a > b . . . 1 and L M is the latus rectum,. So, the equation of latus rectum is: u = x - a e = 0 . . . 2 We know that, when a conic touches a second conic at each of two points, the two conics are said to have double contact with one another. The two conics S = λ u 2 . . . 3 and S = 0 Therefore, have double contact with one another, the straight line u = 0 passing through the two points of contact. So, from equation 3 , the equation of the required circle is: x 2 a 2 + y 2 b 2 - 1 = λ x - a e 2 ⇒ x 2 a 2 + y 2 b 2 - 1 = λ x 2 - 2 a e x + a 2 e 2 ⇒ x 2 1 a 2 - λ + y 2 b 2 + 2 λ a e x - 1 - λ a 2 e 2 = 0 . . . 4 Now, if equation 4 will represent a circle then: 1 a 2 - λ = 1 b 2 ⇒ λ = 1 a 2 - 1 b 2 ⇒ λ = b 2 - a 2 a 2 b 2 ⇒ λ = - e 2 b 2 ( ∵ e 2 = a 2 - b 2 a 2 ) Put the value λ in the equation 4 , we get: x 2 + y 2 + 2 - e 2 b 2 a e b 2 x - b 2 - - e 2 b 2 a 2 e 2 b 2 = 0 ⇒ x 2 + y 2 - 2 a e 3 x - a 2 1 - e 2 + a 2 e 4 = 0 ⇒ x 2 + y 2 - 2 ae 3 x = a 2 1 - e 2 - e 4 . Hence, Proved.

HARD

JEE Main/Advanced

IMPORTANT

Earn 100

A Parallelogram circumscribes the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and two of its angular points are on the lines $x^{2} - h^{2} = 0$ . Prove that the other two are on the conic $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} (1 - \frac{a^{2}}{h^{2}}) = 1$ .

Important Questions on Ellipse

HARD

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 6 - Ellipse>EXERCISE ON LEVEL-II>Q 12

If the normal at

P (θ)

and

Q (\frac{π}{2} + θ)

to the ellipses

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

meet the major axis at

G

and

g

, respectively, then the value of

P G^{2} + Q g^{2}

a^{2} (e^{2} - 1) (e^{2} - 2) .

HARD

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 6 - Ellipse>EXERCISE ON LEVEL-II>Q 13

An ellipse is rotated through a right angle in its own plane about its center, which is fixed; prove that the locus of the point of intersection of a tangent to the ellipse

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

in its original position, with the tangent at the same point of the curve in its new position is

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

MEDIUM

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 6 - Ellipse>EXERCISE ON LEVEL-II>Q 14

Prove that the area of the triangle formed by lines joining the points on the ellipse

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

whose eccentric angles are

α, β, γ

\frac{1}{2} a b \{\sin (β - γ) + \sin (α - β) + \sin (γ - α)\} .

HARD

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 6 - Ellipse>EXERCISE ON LEVEL-II>Q 16

The eccentric angles of two points

P

and

Q

on the ellipse

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

are

α

and

β

. Prove that the area of the parallelogram formed by the tangents at the ends of the diameters through

P

and

Q

\frac{4 a b}{\sin (α - β)}

and hence that it is least when

P

and

Q

are the extremities of a pair of conjugate diameters.

HARD

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 6 - Ellipse>EXERCISE ON LEVEL-II>Q 19

Prove that the directrices of the two parabolas that can be drawn to have their foci at any given point

P

of the ellipse

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

and to pass through its foci meet at an angle which is equal to twice the eccentric angle of

P .

HARD

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 6 - Ellipse>EXERCISE ON LEVEL-II>Q 21

Prove that the equation to the circle, having double contact with the ellipse

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

at the ends of a latus rectum, is

x^{2} + y^{2} - 2 a e^{3} x = a^{2} (1 - e^{2} - e^{4})

MEDIUM

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 6 - Ellipse>EXERCISE ON LEVEL-II>Q 22

Two equal ellipses of eccentricity

e

are placed with their axes at right angles, and have a common focus

S

. If

P Q

be a common tangent, show that the angle

P S Q

2 \sin^{- 1} (e / \sqrt{2})

HARD

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 6 - Ellipse>EXERCISE ON LEVEL-II>Q 23

If two semi-conjugate diameters

C P

and

C Q

of an ellipse cut the director circle in

A

and

B

, prove that the straight line

A B

touches the ellipse.

A Parallelogram circumscribes the ellipse x2a2+y2b2=1 and two of its angular points are on the lines x2-h2=0. Prove that the other two are on the conic x2a2+y2b21-a2h2=1.

Important Questions on Ellipse

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A Parallelogram circumscribes the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and two of its angular points are on the lines $x^{2} - h^{2} = 0$ . Prove that the other two are on the conic $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} (1 - \frac{a^{2}}{h^{2}}) = 1$ .