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