The points on the curve $\frac{x^2}{9}+\frac{y^2}{16}=1$ at which the tangents are parallel to x-axis are:

$(0, \pm 4)$

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 $(0, \pm 4)$