The points on the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$ at which the tangents are parallel to the x-axis are :

(0, ±5)

The correct answer is Option (3) → (0, ±5)

$\frac{x^2}{9}+\frac{y^2}{25}=1$   ....(1)

differentiating wrt x

$\frac{2x}{9}+\frac{2y}{25}\frac{dy}{dx}=0$

so $\frac{2y}{25}\frac{dy}{dx}=\frac{2x}{9}$

$\frac{dy}{dx}=-\frac{25x}{9y}$ for $\frac{dy}{dx}=0$ to be parellel to x axis

so $-\frac{25}{9}\frac{x}{y}=0⇒x=0$

so from (1)

$\frac{y^2}{25}=1⇒y=±5$

points (0, ±5)