Question

If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity $e$ of the ellipse satisfies

• A. $e^4+2e^2-1=0$
• B. $e^2+2e-1=0$
• C. $e^4+e^2-1=0$
• D. $e^2+e-1=0$

Answer 1

The correct option is C

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

The explanation for the correct answer.

Step 1: Find the equation of normal?

Given: Normal at an end of the latus rectum of an ellipse passing through from the end of the minor axis.

Coordinates of the ellipse at the end of the latus rectum is $\left(ae,\frac{b^2}{a}\right)$

The coordinates of the ellipse at the end of the minor axis is $\left(0,-b\right)$

The equation of normal is given by $\frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=a^2{e}^2$

Equation of normal at the end of the latus rectum

$$\begin{gathered} \Rightarrow \frac{a^{2}x}{ae}-\frac{b^{2}y}{{\displaystyle \raisebox{1ex}{$b^2$}\!\left/ \!\raisebox{-1ex}{$a$}\right.}}=a^2{e}^2 \\ \Rightarrow \frac{ax}{e}-ay=a^2{e}^2 \\ \Rightarrow \frac{x}{e}-y=a{e}^2 \end{gathered}$$

Step 2: Put the point $\left(0,-b\right)$in the equation of normal

Normal is passing through $\left(0,-b\right)$

$$\begin{gathered} \Rightarrow 0-(-b)=a{e}^2 \\ \Rightarrow b=a{e}^2 \\ \Rightarrow b^2=a^2{e}^4 \\ \Rightarrow a^2(1-e^2)=a^2{e}^4\left[b^2=a^2(1-e^2)\right] \\ \Rightarrow 1-e^2=e^4 \\ \Rightarrow e^4+e^2-1=0 \end{gathered}$$

Hence option (c) is correct.