Question

The equation of the normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the positive end of latus rectum in the first quadrant

• A. $x+ey+e^{2}a=0$
• B. $x-ey-e^{3}a=0$
• C. $x-ey-e^{2}a=0$
• D. None of these

Answer 1

The correct option is B

$x-ey-e^{3}a=0$

The positive end of latus rectum is $(ae,\frac{b^2}{a})$

$\text{a cos}\theta =\text{ae}$ $\therefore \text{cos}\theta =\text{e}$

$\text{b sin}\theta =\frac{b^2}{a}$ and $\text{sin}\theta =\frac{b}{a}$

Equation of normal is $\frac{ax}{\text{cos}\theta }-\frac{by}{\text{sin}\theta }=a^2-b^2$

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

or $x-ey=a{e}^3$

$\therefore x-ey-e^{3}a=0$