Question

Let 'd' be the perpendicular distance from the centre of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}$= 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) $4\left(1-\frac{b^2}{d^2}\right)$
• (B) $4a^2\left(1-\frac{b^2}{d^2}\right)$Your Answer
• (C) $4b^2\left(1-\frac{a^2}{b^2}\right)$
• (D) None of these

Answer 1

The Correct Option is C

Explanation:
Given:'d' is the perpendicular distance from the centre of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}$ = 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  $\frac{x^2}{a^2}+\frac{y^2}{b^2}= 1$ (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 = $e\left(\frac{a}{e}-x\right)$ and S'P = $e\left(\frac{a}{e}+x\right)$⇒ SP = a - ex and S'P = a + ex

Consider, 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,$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$$\Rightarrow \mathrm{d}=\frac{1}{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}$ [Using perpendicular distance of a line from a point and trigonometric identities ]$\Rightarrow \frac{1}{d^2}=\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}$$\Rightarrow \frac{b^2}{d^2}=\frac{b^2}{a^2}{\cos }^2\theta +{\sin }^2\theta$$\begin{array}{rl}\text{ or }1-\frac{b^2}{d^2}=1& -\frac{b^2}{a^2}{\cos }^2\theta -{\sin }^2\theta ={\cos }^2\theta -\frac{b^2}{a^2}\cos \theta \\ & ={\cos }^2\theta (1-\frac{b^2}{a^2})\end{array}$$={\cos }^2\theta \cdot e^2[\text{ Using eccentricity of ellipse }]$$\Rightarrow 4a^2(1-\frac{b^2}{d^2})=4a^2{\cos }^2\theta \cdot e^2$So, ${({PF}_1-{PF}_2)}^2=4a^2{e}^2{\cos }^2\theta =4a^2(1-\frac{b^2}{d^2})[\mathrm{Using}(i)]$Hence, the correct option is (B).