The points on the curve $\frac{x^2}{9}+\frac{y^2}{64}= 1$ at which the tangents are parallel to the x-axis are :
Answer & explanation
Correct answer: option 3
$\frac{x^2}{9}+\frac{y^2}{64}=1.$
$\text{Differentiate implicitly:}$
$\frac{2x}{9}+\frac{2y}{64}\frac{dy}{dx}=0.$
$\frac{dy}{dx}=-\frac{64x}{9y}.$
$\text{For tangents parallel to x-axis:}$
$\frac{dy}{dx}=0.$
$-\frac{64x}{9y}=0.$
$x=0.$
$\text{Substitute in ellipse equation:}$
$\frac{y^2}{64}=1.$
$y=\pm8.$
$\text{Points }=(0,8),(0,-8).$