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

Let the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a>b$. The foci are $F_1(-c,0)$ and $F_2(c,0)$, where $c^2 = a^2(1-e^2)$. 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 $P(0, \pm a)$. Let's take $P(0, a)$. The orthocentre $H$ of $\triangle PF_1F_2$ with $P(0, a)$, $F_1(-c, 0)$, $F_2(c, 0)$ is found to be $H(0, \frac{c^2}{a})$. Since $H$ lies on the ellipse, we substitute its coordinates into the ellipse equation: $\frac{0^2}{a^2} + \frac{(c^2/a)^2}{b^2} = 1 \implies \frac{c^4}{a^2 b^2} = 1 \implies c^4 = a^2 b^2$. Using $c^2 = a^2e^2$ and $b^2 = a^2(1-e^2)$: $(a^2e^2)^2 = a^2 (a^2(1-e^2))$ $a^4e^4 = a^4(1-e^2)$ $e^4 = 1-e^2$. Let $x = e^2$. Then $x^2 = 1-x \implies x^2+x-1=0$. Solving for $x$: $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$. Since $e^2 > 0$, we take $e^2 = \frac{-1+\sqrt{5}}{2}$. We know that $\sin 18^\circ = \frac{\sqrt{5}-1}{4}$. Therefore, $2\sin 18^\circ = 2 \left(\frac{\sqrt{5}-1}{4}\right) = \frac{\sqrt{5}-1}{2}$. Thus, $e^2 = 2\sin 18^\circ$.