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 : |
(±5, 0) (0, ±3) (0, ±5) (±3, 0) |
(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) |