The points on the curve $\frac{x^2}{9}+\frac{y^2}{16}=1$ at which the tangents are parallel to x-axis are:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) - $(0, \pm 4)$
tangents are parallel to x-axis
$⇒\frac{dy}{dx}=0$
differentiating equation w.r.t. x
$\frac{2x}{9}+\frac{2y}{16}\frac{dy}{dx}=0$ as $\frac{dy}{dx}=0$
$⇒\frac{2x}{9}=0⇒x=0$
from equation of curve at $x=0$
$\frac{y^2}{16}=1⇒y=±4$
Point $(0, \pm 4)$