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 F₂ are the two foci of the ellipse then
(PF₁ – PF₂)² = K $\left( 1 - \frac{b^{2}}{d^{2}} \right)$where K=

• A. 4a²
• B. 3a²
• C. 2a²
• D. None of these

Answer 1

Answer: (A)

4a²

Answer 2

Let tangent is x = a and P(a, 0)

Now, PF₁ = a – ae, PF₂ = a + ae ̃ d = a

= (PF₁ – PF₂)² ̃ (a – ae – (a + ae))² = 4a²e²

= 4a²$\left( 1 - \frac{b^{2}}{a^{2}} \right)$ = 4a²$\left( 1 - \frac{b^{2}}{d^{2}} \right)$