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

(0, 4), (0, -4)

The correct answer is Option (1) → (0, 4), (0, -4)

$\frac{x^2}{9}+\frac{y^2}{16}=1$

tangents parallel to x axis ⇒ $\frac{dy}{dx}=0$

so differentiating wrt x

so $\frac{2x}{9}+\frac{2y}{16}\frac{dy}{dx}=0$  so x = 0

so $\frac{(0)^2}{9}+\frac{y^2}{16}=1⇒y=±4$

points (0, 4), (0, -4)