Question

If the circle whose diameter is the major axis of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \, \left(a > b > 0\right)$ meets the minor axis at point $P$ and the orthocentre of $\Delta PF_{1}F_{2}$ lies on the ellipse, where $F_{1}$ and $F_{2}$ are the foci of the ellipse, then the square of the eccentricity of the ellipse is

• A. $\frac{\sqrt{5} - 1}{2}$
• B. $\sqrt{3}-1$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

Answer: (A)

$m_{F_{1} B}\cdot m_{P F_{2}}=-1\Rightarrow \frac{a - 0}{0 - a e}\times \frac{b - 0}{0 - \left(- a e\right)}=-1\Rightarrow \frac{b^{2}}{a^{2}}=e^{2}\Rightarrow e^{4}+e^{2}-1=0$
$\therefore e^{2}=\left(\frac{\sqrt{5} - 1}{2}\right)$