If the eccentric angles of the ends of a focal chord of the ellipse x 2 a 2 + y 2 b 2 = 1 be θ and ψ , show that tan θ 2 tan ψ 2 = e - 1 e + 1 .

Equation of line joining points θ , ψ is x a cos θ + ψ 2 + y b sin θ + ψ 2 = cos θ - ψ 2 If it is a focal chord, then it passes through focus a e , 0 , so e cos θ + ψ 2 = cos θ - ψ 2 ⇒ cos θ - ψ 2 cos θ + ψ 2 = e ⇒ cos θ - ψ 2 - cos θ + ψ 2 cos θ + ψ 2 + cos θ - ψ 2 = e - 1 e + 1 ⇒ 2 sin θ 2 sin ψ 2 2 cos θ 2 cos ψ 2 = e - 1 e + 1 ⇒ tan θ 2 tan ψ 2 = e - 1 e + 1 Now using - a e , 0 , we get tan θ 2 tan ψ 2 = e + 1 e - 1

Show that each pair of tangents at the ends of the latus rectum of the ellipse x 2 a 2 + y 2 b 2 = 1 passes through the foot of the corresponding directrix.

Given E = x 2 a 2 + y 2 b 2 = 1 Ends of latus rectum = a e , b 2 a , a e , - b 2 a Tangent to ellipse is T = x x 1 a 2 + y y 1 b 2 = 1 Let L 1 = a e , b 2 a , L 2 = a e , - b 2 a T 1 = e x a + y a = 1 . . . . . 1 T 2 = e x a - y a = 1 . . . . . 2 Add 1 and 2 , x = a e , y = 0 ( a e , 0 ) this is the point of intersection.

Show that the tangents at the extremities of all chords of the ellipse x 2 a 2 + y 2 b 2 = 1 which subtend a right angle at the centre, intersect on the ellipse x 2 a 4 + y 2 b 4 = 1 a 2 + 1 b 2

Given ellipse x 2 a 2 + y 2 b 2 = 1 , Let P = ( a 2 + b 2 cos q , a 2 + b 2 sin q ) Equation of chord of contact of tangents from P is x a 2 a 2 + b 2 cos q + y b 2 a 2 + b 2 sin q = 1 Let this line be a tangent to the fixed ellipse x 2 A 2 + y 2 B 2 = 1 A = a 2 a 2 + b 2 , B = b 2 a 2 + b 2 Hence ellipse is x 2 a 4 + y 2 b 4 = 1 a 2 + b 2 .