If a tangent to the ellipse x 2 a 2 + y 2 b 2 = 1 ( a > b ) meets its major axis and minor axis at M and N respectively, then prove that a 2 ( C M ) 2 + b 2 ( C N ) 2 = 1 , where C is the centre of the ellipse.

Let P ( θ ) = ( a cos ⁡ θ , b sin ⁡ θ ) be a point on the ellipse x 2 a 2 + y 2 b 2 = 1 . Then, the equation of the tangent at P ( θ ) is x cos ⁡ θ a + y sin ⁡ θ b = 1 i.e. x a cos θ + y b sin θ = 1 meets major axis ( x - axis) and minor axis ( y - axis) at M and N respectively. ∴ C M = a cos ⁡ θ , C N = b sin ⁡ θ ⇒ a C M = cos ⁡ θ , b C N = sin ⁡ θ Squaring both sides, we get: a 2 C M 2 = cos 2 ⁡ θ . . . 1 , b 2 C N 2 = sin 2 ⁡ θ . . . 2 Adding 1 and 2 , we get: ∴ a 2 ( C M ) 2 + b 2 ( C N ) 2 = cos 2 ⁡ θ + sin 2 ⁡ θ = 1 ∵ cos 2 ⁡ z + sin 2 ⁡ z = 1 Hence proved.

Find the equation of tangents to the ellipse 2 x 2 + y 2 = 8 which is perpendicular to x + y + 2 = 0

Equation of ellipse 2 x 2 + y 2 = 8 ⇒ x 2 4 + y 2 8 = 1 On comparing with standard equation of ellipse, we get Here a 2 = 4 , b 2 = 8 Given equation is x + y + 2 = 0 ⇒ y = - x - 2 , slope is m 1 = - 1 Since the tangent is perpendicular to the given equation then the slope is m 2 = 1 Therefore, equation of tangent is y = x + c If y = m x + c is a tangent to the ellipse then c = a 2 m 2 + b 2 ⇒ c = ± 4 1 + 8 = ± 2 3 Hence, equation of tangent to ellipse 2 x 2 + y 2 = 8 is y = x ± 2 3

Find the equation of tangents to the ellipse 2 x 2 + y 2 = 8 which is parallel to x - 2 y + 4 = 0 .

Equation of the ellipse 2 x 2 + y 2 = 8 ⇒ x 2 4 + y 2 8 = 1 Equation of the tangent parallel to x - 2 y + 4 = 0 is x - 2 y + k = 0 ⇒ y = x 2 + k 2 ⇒ m = 1 2 and c = k 2 Above line is a tangent to the ellipse ⇒ c 2 = a 2 m 2 + b 2 ⇒ k 2 4 = 4 · 1 4 + 8 ⇒ k 2 = 36 ⇒ k = ± 6 Equations of tangents are x - 2 y ± 6 = 0 .

Find the equation of the ellipse referred to its major and minor axes as the co-ordinate axes x , y respectively with latus rectum of length 4 and distance between foci 4 2 .

Let the equation of the ellipse be x 2 a 2 + y 2 b 2 = 1 a > b . Length of latus rectum = 2 b 2 a = 4 ⇒ b 2 = 2 a Distance between foci S = a e , 0 , S ' = - a e , 0 is 2 a e = 4 2 ⇒ a e = 2 2 But b 2 = a 2 1 − e 2 ⇒ 2 a = a 2 − ( a e ) 2 = a 2 − 8 ⇒ a 2 − 2 a − 8 = 0 ⇒ ( a − 4 ) ( a + 2 ) = 0 ⇒ a = 4 ( ∵ a > 0 ) So, b 2 = 2 a = 8 Thus, the equation of ellipse x 2 16 + y 2 8 = 1 ( or ) x 2 + 2 y 2 = 16

Find the equation of the ellipse in the standard form whose distance between foci is 2 and length of latus rectum is 15 2 .

Given distance between foci = 2 a e = 2 ⇒ a e = 1 And length of latus rectum = 2 b 2 a = 15 2 ⇒ b 2 a 2 = 15 4 a ⇒ b 2 a 2 = 15 e 4 ∵ a e = 1 Now, we have e 2 = 1 - b 2 a 2 = 1 - 15 e 4 ⇒ 4 e 2 + 15 e - 4 = 0 Using quadratic formula, we get e = - 15 ± 225 + 64 8 ⇒ e = - 15 ± 17 8 ⇒ e = 1 4 Therefore, a = 4 and b = 15 Hence, standard equation of ellipse is given by x 2 16 + y 2 15 = 1 .