Question

If normal at one extremity of latus rectum of an ellipse passes through one extremity of minor axis, prove that the eccentricity e is given by e^4+e^2,-1=0

Answer 1

Let the equation of the ellipse is:

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

The equation of the normal to this ellipse at $\left(ae , \dfrac{b^2}{a}\right)$ is:

$$\dfrac{x-ae}{\frac{ae}{a^2}} = \dfrac{y-\frac{b^2}{a}}{\frac{b^2}{ab^2}} \ \rightarrow (1) \\\\$$

This normal passes through $(0,−b)$  according to given condition.

Substitute $x=0 , y=-b$ in equation (1)

$$\dfrac{0-ae}{\frac{ae}{a^2}} = \dfrac{-b-\frac{b^2}{a}}{\frac{b^2}{ab^2}}$$

Simplify

$$-a^2=-ab-b^2 \ \rightarrow (2)$$

Since $e^2=1-\dfrac{b^2}{a^2}$

So, $\dfrac{b^2}{a^2}= 1-e^2$

Therefore, $b^2=a^2(1-e^2)$

Substitute $b^2$ in equation (2)

$$-a^2=-ab-a^2(1-e^2)$$

Simplify

$$a^2=ab+a^2-a^2e^2$$

$$ab=a^2e^2$$

$$b=ae^2$$

So,

$$b^2 = a^2e^4$$

And we know that

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

Therefore,

$$a^2e^4=a^2(1-e^2)$$

Simplify

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