Question

If the normal at the end of a latus rectum of an ellipse passes through one extremity of the minor axis, then

• A. $e^4+e^2-1=0$
• B. $e^4-e^2+1=0$
• C. $e^4-e^2-1=0$
• D. None of these

Answer 1

The correct option is A $e^4+e^2-1=0$

Let the equation of the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Let the normal at the extremity $L$ of the latus rectum passes though the extremity $B^{\prime }$ of the minor axis.

Coordinates of $L$ are $(ae,\frac{b^2}{a})$ and coordinates of $B^{\prime }$ are $(0,-b)$

Equation of the normal at $L$ is

$\frac{a^2.x}{ae}-\frac{b^2.y}{b^2/a}=a^2-b^2$ ........ $[\because \frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=a^2-b^2]$

$\Rightarrow \frac{ax}{e}-ay=a^2-b^2$

If it passes through $B^{\prime }(0,-b)$, then $0+ab=a^2-b^2\Rightarrow a^2{b}^2={(a^2-b^2)}^2$

But $b^2=a^2(1-e^2)$

$\therefore a^2\times a^2(1-e^2)={[a^2-a^2(1-e^2)]}^2\Rightarrow a^4(1-e^2)=a^4{(1-1+e^2)}^2\Rightarrow 1-e^2=e^4\Rightarrow e^4+e^2-1=0$