Question

If circle whose diameter is major axis of ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (a > b) meets minor axis at point P & orthocentre of $\triangle PF_1F_2$ lies on ellipse where $F_1$ & $F_2$ are focii of ellipse, then square of eccentricity of ellipse is-

• A. 2 sin18°
• B. 2 sin15°
• C. sin45°
• D. sin60°

Answer 1

Answer: (A)

2 sin18°

Answer 2

The ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The foci are $F_1(-ae, 0)$ and $F_2(ae, 0)$. The circle with the major axis as diameter has equation $x^2 + y^2 = a^2$. This circle meets the minor axis (y-axis) at $P(0, a)$. The orthocentre of $\triangle PF_1F_2$ is $H(0, ae^2)$. For H to lie on the ellipse, we have $\frac{(ae^2)^2}{b^2} = 1$, which implies $a^2e^4 = b^2$. Using $b^2 = a^2(1-e^2)$, we get $e^4 = 1-e^2$, so $e^4 + e^2 - 1 = 0$. Solving for $e^2$ gives $e^2 = \frac{-1+\sqrt{5}}{2}$. We know that $\sin 18^\circ = \frac{\sqrt{5}-1}{4}$, so $2\sin 18^\circ = \frac{\sqrt{5}-1}{2}$. Thus, $e^2 = 2\sin 18^\circ$.