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
Answer & explanation
Correct answer: option 1
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 $(0,-3)$.