Question

The tangent at a point P (acosθ, bsinθ) on the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ meets the auxiliary circle in two points. The chord joining them subtends a right angle at the center. Then the eccentricity of the ellipse is given by

• A. (1 + sin²θ)^-1/2
• B. (1+cos²θ)^-1/2
• C. (1+sin²θ)
• D. (1+cos²θ)^1/2

Answer 1

Answer: (A)

(1 + sin²θ)^-1/2

Answer 2

$x^{2} + y^{2} = a^{2}\left( \frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta \right)$The lines given by this equation are at right angles.

∴ coefficient of x² + coefficient of y² = 0

⇒$1 - a^{2}\left( \frac{\cos^{2}\theta}{a^{2}} \right) + 1 - a^{2}\left( \frac{\sin^{2}\theta}{b^{2}} \right) = 0$⇒ sin²θ + 1-

$\frac{a^{2}}{b^{2}}\sin^{2}\theta = 0$

⇒ sin²θ + $\left( 1 - \frac{a^{2}}{b^{2}} \right) + 1 = 0$

⇒ = sin²θ (a²-b²)+b² = 0

⇒ $- a^{2}e^{2}\sin^{2}\theta + a^{2}\left( 1 - e^{2} \right) = 0$⇒ 1 = e² (1+sin²θ)

⇒ e = (1+sin²θ)^-1/2.