The point on the curve $\frac{x^2}{4} +\frac{y^2}{9}= 1$ at which the tangent to the curve is parallel to the x-axis is

(0, 3)

The correct answer is Option (1) → (0, 3) **

$\frac{x^2}{4}+\frac{y^2}{9}=1$

Differentiate:

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

$\frac{dy}{dx}=-\frac{9x}{4y}$

For tangent ∥ x-axis:

$\frac{dy}{dx}=0\;$ gives $\;x=0$

Substitute $x=0$ into ellipse:

$\frac{y^2}{9}=1\;\Rightarrow\;y=\pm 3$

The required points are $(0,3)$ and $(0,-3)$.