Question

Let 'd' be the perpendicular distance from the centre of the ellipse x2a2+y2b2= 1 to the tangent drawn at a point P on the ellipse. If F1 and F are two foci of the ellipse, then (PF1 - PF)2 is equal to;

• A. (A) 41-b2d2
• B. (B) 4a21-b2d2Your Answer
• C. (C) 4b21-a2b2
• D. (D) None of these

Answer 1

Answer: (C)

(C) 4b21-a2b2

Answer 2

Explanation:
Given:'d' is the perpendicular distance from the centre of the ellipse x2a2+y2b2 = 1 to the tangent drawn at a point P on the ellipse.-- F1 and F are two foci of the ellipse.We have to find the value of (PF1 - PF)2.Consider,Let P(x, y) be any point on the ellipse x2a2+y2b2= 1 (As shown in Fig).Then, by definition of ellipse, we haveSP = e PM and S'P = e PM'⇒ S = e(NK) and S'P = e(NK')⇒ SP = e(CK - CN) and S'P = e(CK' + CN)⇒ SP = eae-x and S'P = eae+x⇒ SP = a - ex and S'P = a + exConsider, co-ordinate of point P in parametric form as (a cos θ, b sin θ).Since PF = a + ex and PF = a - ex , thereforePF - PF = 2ex = 2ea cos θ∴ (PF - PF)2 = 4a22cos2θ ......... (i)Equation of tangent to ellipse at P(θ) is,xacosθ+ybsinθ=1⇒d=1cos2θa2+sin2θb2 [Using perpendicular distance of a line from a point and trigonometric identities ]⇒1d2=cos2⁡θa2+sin2⁡θb2⇒b2d2=b2a2cos2⁡θ+sin2⁡θ or 1−b2d2=1−b2a2cos2⁡θ−sin2⁡θ=cos2⁡θ−b2 a2cos⁡θ=cos2⁡θ(1−b2a2)=cos2⁡θ⋅e2[ Using eccentricity of ellipse ]⇒4a2(1−b2d2)=4a2cos2⁡θ⋅e2So, (PF1−PF2)2=4a2e2cos2⁡θ=4a2(1−b2d2)[Using(i)]Hence, the correct option is (B).