If the normal at point P θ on the ellipse x 2 a 2 + y 2 b 2 = 1 meets the axes of x and y at M and N respectively, show that P M : P N = b 2 : a 2 .

Given equation of ellipse is x 2 a 2 + y 2 b 2 = 1 Let P ( θ ) = ( a cos θ , b sin θ ) The equation of normal to the given ellipse at P is given by a x cos θ - b y sin θ = a 2 - b 2 Putting y = 0 , we get x - intercept = ( a 2 - b 2 ) cos θ a And putting x = 0 , we get y - intercept = ( b 2 - a 2 ) b sin θ M = ( a 2 - b 2 ) cos θ a , 0 N = 0 , ( b 2 - a 2 ) b sin θ Now, using distance formula we get P M = ( a 2 - b 2 ) cos θ - a 2 cos θ a 2 + b 2 sin 2 θ ⇒ P M = b 4 cos 2 θ a 2 + b 2 sin 2 θ ⇒ P M = b a b 2 cos 2 θ + a 2 sin 2 θ And similarly, P N = a b b 2 cos 2 θ + a 2 sin 2 θ ⇒ P M : P N = b 2 : a 2

If the normal at point P θ on the ellipse x 2 a 2 + y 2 b 2 = 1 meets the axes of x and y at M and N respectively, show that P M : P N = b 2 : a 2 .

Given equation of ellipse is x 2 a 2 + y 2 b 2 = 1 Let P ( θ ) = ( a cos θ , b sin θ ) The equation of normal to the given ellipse at P is given by a x cos θ - b y sin θ = a 2 - b 2 Putting y = 0 , we get x - intercept = ( a 2 - b 2 ) cos θ a And putting x = 0 , we get y - intercept = ( b 2 - a 2 ) b sin θ M = ( a 2 - b 2 ) cos θ a , 0 N = 0 , ( b 2 - a 2 ) b sin θ Now, using distance formula we get P M = ( a 2 - b 2 ) cos θ - a 2 cos θ a 2 + b 2 sin 2 θ ⇒ P M = b 4 cos 2 θ a 2 + b 2 sin 2 θ ⇒ P M = b a b 2 cos 2 θ + a 2 sin 2 θ And similarly, P N = a b b 2 cos 2 θ + a 2 sin 2 θ ⇒ P M : P N = b 2 : a 2

If the normal at the point P ( θ ) to the ellipse x 2 14 + y 2 5 = 1 , intersects it again at the point Q ( 2 θ ) show that cosθ = - 2 3

Given equation of ellipse is x 2 14 + y 2 5 = 1 Here, a 2 = 14 , b 2 = 5 Let P ( θ ) = a cos θ , b sin θ Now, equation of normal at point P ( θ ) is given by 14 x cosθ - 5 y sinθ = 9 . . . . ( 1 ) Now, let Q ( 2 θ ) = a cos 2 θ , b sin 2 θ As equation ( 1 ) passes through Q ( 2 θ ) , ⇒ 14 . 14 cos 2 θ cosθ - 5 . 5 sin 2 θ sinθ = 9 ⇒ 14 cos 2 θ cosθ - 10 cosθ = 9 ⇒ 14 cos 2 θ - 10 cos 2 θ = 9 cosθ ⇒ 14 2 cos 2 θ - 1 - 10 cos 2 θ = 9 cosθ ⇒ 18 cos 2 θ - 9 cosθ - 14 = 0 ⇒ 18 cos 2 θ - 21 cosθ + 12 cosθ - 14 = 0 ⇒ 3 cosθ ( 6 cosθ - 7 ) + 2 ( 6 cosθ - 7 ) = 0 ⇒ cosθ = - 2 3

If the chords through points θ and ϕ on the ellipse x 2 a 2 + y 2 b 2 = 1 intersect the major axis at a distance c from the centre, prove that tan θ 2 tan ϕ 2 = c - a c + a

Given ellipse is x 2 a 2 + y 2 b 2 = 1 with centre at origin Let the point be P ( θ ) = a cos θ , b sin θ and Q ( ϕ ) = a cos ϕ , b sin ϕ . Now, equation of the chord P Q is y - b sin θ = b sin ϕ - b sin θ a cos ϕ - a cos θ x - a cos θ This intersects the major axis at c , 0 0 - b sin θ = b sin ϕ - b sin θ a cos ϕ - a cos θ c - a cos θ ⇒ c = - a sin θ cos ϕ - cos θ sin ϕ - sin θ + a cos θ ⇒ c = 2 a sin θ sin ϕ + θ 2 sin ϕ - θ 2 2 cos ϕ + θ 2 sin ϕ - θ 2 + a cos θ ⇒ c = a sin θ sin ϕ + θ 2 + a cos θ cos ϕ + θ 2 cos ϕ + θ 2 ⇒ c a = cos θ - ϕ + θ 2 cos ϕ + θ 2 ⇒ c a = cos θ - ϕ 2 cos ϕ + θ 2 Now, applying componendo dividendo, we get ⇒ c - a c + a = cos θ - ϕ 2 - cos ϕ + θ 2 cos θ - ϕ 2 + cos ϕ + θ 2 ⇒ c - a c + a = - 2 sin θ 2 sin - ϕ 2 2 cos θ 2 cos - ϕ 2 ⇒ c - a c + a = sin θ 2 sin ϕ 2 cos θ 2 cos ϕ 2 ⇒ c - a c + a = tan θ 2 tan ϕ 2 Hence the result

Prove that the length of the focal chord of the ellipse x 2 a 2 + y 2 b 2 = 1 which is inclined to the major axis at angle θ is 2 a b 2 a 2 sin 2 θ + b 2 cos 2 θ .

Given that, the equation of the ellipse is x 2 a 2 + y 2 b 2 = 1 Let P S Q be the focal chord of the ellipse Therefore, S = a e , 0 and S ' = - a e , 0 , where e is eccentricity of the ellipse Now, the slope of the focal chord = tan θ So, the equation of P S Q is y - 0 = tan θ ( x - a e ) ⇒ y = tan θ ( x - a e ) Let P x 1 y 1 and Q x 2 , y 2 are points on the ellipse Therefore, x 2 a 2 + tan 2 θ ( x - a e ) 2 b 2 = 1 ⇒ b 2 x 2 + a 2 tan 2 θ ( x - a e ) 2 = a 2 b 2 ⇒ b 2 x 2 + a 2 tan 2 θ x 2 - 2 a e x + a 2 e 2 - a 2 b 2 = 0 ⇒ b 2 x 2 + a 2 tan 2 θ x 2 - 2 a 3 e tan 2 θ x + a 4 e 2 tan 2 θ - a 2 b 2 = 0 ⇒ b 2 + a 2 tan 2 θ x 2 - 2 a 3 e tan 2 θ x + a 4 e 2 tan 2 θ - a 2 b 2 = 0 → 1 It is a quadratic equation in x , therefore, sum of roots x 1 + x 2 = 2 a 3 e tan 2 θ b 2 + a 2 tan 2 θ and product of roots x 1 x 2 = a 4 e 2 tan θ - a 2 b 2 b 2 + a 2 tan θ [Assume that, x 1 , x 2 are the roots of equation 1 ] Now, y = tan θ [ x - a e ] So, y 1 = tan θ x 1 - a e and y 2 = tan θ x 2 - a e Thus, y 1 - y 2 = tan θ x 1 - a e - tan θ x 2 - a e = tan θ x 1 - a e - x 2 + a e = tan θ x 1 - x 2 Therefore, the length of focal chord P Q = x 1 - x 2 2 + y 1 - y 2 2 = x 1 - x 2 2 + tan θ x 1 - x 2 2 = x 1 - x 2 ) 2 + tan 2 θ x 1 - x 2 2 = x 1 - x 2 2 1 + tan 2 θ = x 1 - x 2 2 · sec 2 θ = x 1 - x 2 ( sec θ ) = x 1 + x 2 2 - 2 x 1 x 2 · sec θ = 2 a 3 e tan 2 θ b 2 + a 2 tan 2 θ 2 - 4 × a 4 e 2 tan θ - a 2 b 2 b 2 + a 2 tan θ · sec θ = 4 a 6 e 2 tan 4 θ b 2 + a 2 + tan 2 θ 2 - 4 a 4 e 2 tan 2 θ - 4 a 2 b 2 b 2 + a 2 tan 2 θ · sec θ = 4 a 6 e 2 tan 4 θ - 4 a 4 e 2 tan 2 θ - 4 a 2 b 2 b 2 + a 2 tan 2 θ b 2 + a 2 tan 2 θ 2 · sec θ = 4 a 6 e 2 tan 4 θ - 4 a 4 b 2 e 2 tan 2 θ + 4 a 6 e 2 tan 4 θ - 4 a 2 b 4 - 4 a 4 b 2 + 4 a 4 b 2 tan 2 θ · sec θ b 2 + a 2 tan 2 θ = 4 a 2 b 4 + 4 a 4 b 2 tan 2 θ - 4 a 4 b 2 e 2 tan 2 θ · sec θ b 2 + a 2 tan 2 θ = 4 a 2 b 2 b 2 + a 2 tan 2 θ - a 2 e 2 tan 2 θ · sec θ b 2 + a 2 tan 2 θ = b 2 + a 2 tan 2 θ - a 2 e 2 tan 2 θ · 2 a b sec θ b 2 + a 2 tan 2 θ = b 2 + a 2 tan 2 θ - a 2 a 2 - b 2 a 2 tan 2 θ · 2 a b sec θ b 2 + a 2 tan 2 θ [Using e = 1 - b 2 a 2 = a 2 - b 2 a 2 ] = b 2 + a 2 tan 2 θ - a 2 tan 2 θ + b 2 tan 2 θ · 2 a b secθ b 2 + a 2 tan θ = b 2 sec 2 θ · 2 a b sec θ b 2 + a 2 tan 2 θ = b sec θ · 2 a b sec θ b 2 + a 2 tan 2 θ = 2 a b 2 sec 2 θ b 2 + a 2 tan 2 θ Now, dividing numerator and denominator with sec 2 θ = 2 a b 2 a 2 sin 2 θ + b 2 cos 2 θ Hence, proved.

Find the length of the chord of the ellipse x 2 25 + y 2 16 = 1 bisected at the point 1 2 , 2 5 .

Given equation of ellipse is = x 2 25 + y 2 16 = 1 . . . . 1 We know that the equation of chord of an ellipse where mid-point is x 1 , y 1 is given by = x x 1 25 + y y 1 16 - 1 = x 1 2 25 + y 1 2 16 - 1 Since x 1 , y 1 = 1 2 , 2 5 ∴ We have = x 1 2 25 + y 2 5 16 = 1 2 2 25 + 2 5 2 16 ⇒ x 50 + y 40 = 1 100 + 1 100 ⇒ x 50 + y 40 = 2 100 ⇒ x 5 + y 4 = 1 5 or 4 x + 5 y = 4 . . . . 2 From equation 1 x 2 25 + y 2 16 = 1 ⇒ 16 x 2 + 25 y 2 = 400 ⇒ 16 x 2 + 4 - 4 x 2 = 400 From 2 , 5 y = 4 - 4 x ⇒ 16 x 2 + 16 + 16 x 2 - 32 x = 400 ⇒ 32 x 2 - 32 x - 384 = 0 ⇒ x 2 - x - 12 = 0 ⇒ x - 4 x + 3 = 0 ⇒ x = 4 , - 3 If x = 4 , 5 y = 4 - 4 · 4 And if x = - 3 , 5 y = 4 - 4 - 3 ⇒ y = - 12 5 ⇒ y = 16 5 ∴ The points of intersection of chord and ellipse are 4 , - 12 5 & - 3 , 16 5 Length of chord = - 3 - 4 2 + 16 5 + 12 5 2 = - 7 2 + 28 5 2 = 49 + 784 25 = 1225 + 784 25 = 2009 25 = 7 41 5

Show that the length of the chord intercepted by the ellipse x 2 a 2 + y 2 b 2 = 1 on the line y = m x + c is 2 a b a 2 m 2 + b 2 1 + m 2 a 2 m 2 + b 2 - c 2 .

Given line is y = m x + c and ellipse is x 2 a 2 + y 2 b 2 = 1 For the point of intersection of the above two equations, we put value of y = m x + c in equation of ellipse, we get x 2 a 2 + m x + c 2 b 2 = 1 ⇒ b 2 x 2 + a 2 m x + c 2 a 2 b 2 = 1 ⇒ b 2 x 2 + a 2 m 2 x 2 + a 2 c 2 + 2 m c a 2 x = a 2 b 2 ⇒ b 2 + a 2 m 2 x 2 + 2 m c a 2 x + a 2 c 2 - b 2 = 0 Let the two roots of this equation be x 1 , x 2 Then, x 1 + x 2 = - 2 m c a 2 b 2 + a 2 m 2 and x 1 x 2 = a 2 c 2 - b 2 b 2 + a 2 m 2 Now, we know x 2 - x 1 2 = x 1 + x 2 2 - 4 x 1 x 2 ⇒ x 2 - x 1 2 = - 2 m c a 2 b 2 + a 2 m 2 2 - 4 a 2 c 2 - b 2 b 2 + a 2 m 2 ⇒ x 2 - x 1 2 = 4 m 2 c 2 a 4 b 2 + a 2 m 2 2 - 4 a 2 c 2 - b 2 b 2 + a 2 m 2 ⇒ x 2 - x 1 2 = 4 a 2 b 2 + a 2 m 2 2 m 2 c 2 a 2 - c 2 - b 2 b 2 + a 2 m 2 ⇒ x 2 - x 1 2 = 4 a 2 b 2 + a 2 m 2 2 m 2 a 2 b 2 - c 2 b 2 + b 4 ⇒ x 2 - x 1 2 = 4 a 2 b 2 b 2 + a 2 m 2 2 m 2 a 2 - c 2 + b 2 ⇒ x 2 - x 1 = 2 a b b 2 + a 2 m 2 m 2 a 2 - c 2 + b 2 Now, let the end points of chord be x 1 , m x 1 + c and x 2 , m x 2 + c ⇒ Length of chord = x 2 - x 1 2 + m x 2 + c - m x 1 - c 2 = x 2 - x 1 2 + m 2 x 2 - x 1 2 = x 2 - x 1 1 + m 2 = 2 a b b 2 + a 2 m 2 m 2 a 2 - c 2 + b 2 × 1 + m 2 = 2 a b b 2 + a 2 m 2 1 + m 2 m 2 a 2 - c 2 + b 2 Hence, the result