Question

If the normal at the end of latus rectum of the ellipse $\frac{x^2}{a^2}+y^2{b}^2=1$ passes through an extremity of minor axis, then $e^4+e^2-1=0$ or $e^2=\frac{\sqrt{5-1}}{2}$. A

• A. True
• B. False

Answer 1

The correct option is A True

The equation of the ellipse is given by,

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Coordinates of one end of latusrectum is $A(ae,\frac{b^2}{a})$

The normal through point A is passing through one end of minor axis as shown in figure with coordinates $(0,-b)$

The equation of normal from any point on the ellipse is given by,

$\frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=c^2$

$\therefore \frac{a^{2}x}{ae}-\frac{b^{2}y}{\left(\frac{b^2}{a}\right)}=a^2-b^2$

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

This normal is passing through $(0,-b)$. Thus, equation of normal becomes,

$\therefore \frac{a\left(0\right)}{e}-a(-b)=a^2-b^2$

$\therefore ab=a^2-b^2$

Dividing both sides by $a^2$, we get,

$\frac{b}{a}=1-\frac{b^2}{a^2}$

But, $1-\frac{b^2}{a^2}=e^2$

$\therefore \frac{b}{a}=e^2$

Squaring both sides, we get,

$\frac{b^2}{a^2}=e^4$

$\therefore 1-e^2=e^4$

$\therefore e^4+e^2-1=0$