Question

Equation of the tangent to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a^2>b^2)$ at the end of latus rectum in the first quadrant is:

• A. $ax+ey-a=0$
• B. $ex+y-a=0$
• C. $x-ey+a=0$
• D. $ex-2y+a=0$

Answer 1

Answer: (B)

$ex+y-a=0$

End of latus rectum is $(ae,\frac{b^2}{a})$
$\therefore$Tangent at $(ae,\frac{b^2}{a})$ is $\frac{x\left(ae\right)}{a^2}+\frac{y\left(\frac{b^2}{a}\right)}{b^2}=1$
$\frac{xe}{a}+\frac{y}{a}=1$
$ex+y-a=0$ is equation of tangent