Question

The equation of the normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the end of latus rectum in quadrant $1^{st}$ and $4^{th}$ is

• A. $x-ey-a{e}^3=0$
• B. $x+ey-a{e}^3=0$
• C. $y-ex-b{e}^3=0$
• D. $y+ex-b{e}^3=0$

Answer 1

Answer: (A), (B)

The correct options are
A $x-ey-a{e}^3=0$
B $x+ey-a{e}^3=0$

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

ends of L.R in 1st and 4th quadrant is $L(ae,\frac{b^2}{a})$ and $L^{\prime }(ae,\frac{-b^2}{a})$

equation of normal at $L$ and $L^{\prime }$

at $L$, $\Rightarrow \frac{a^{2}x}{ae}-\frac{b^{2}y}{\frac{b^2}{a}}=a^2-b^2$

$\Rightarrow ax-aey=e(a^2-b^2)$

$\Rightarrow x-ey=\frac{a^2}{a}\left(e^2\right)\left(e\right)$

$\Rightarrow x-ey-a{e}^3=0---\left(i\right)$

at $L^{\prime }$ $\Rightarrow \frac{a^{2}x}{ae}-\frac{b^{2}y}{-\frac{b^2}{a}}=a^2{e}^2$

$\Rightarrow x+ey-a{e}^3=0---\left(ii\right)$