The points on the curve (x^2)/9+(y^2)/16=1 at which the tangents are…

The correct answer is Option (1) → (0, 4), (0, −4)

A tangent is parallel to the x-axis where its slope dy/dx is 0.

Differentiating the equation of the curve with respect to x:

    \( \displaystyle \frac{2x}{9} + \frac{2y}{16}\,\frac{dy}{dx} = 0 \)

    \( \displaystyle \frac{dy}{dx} = -\frac{16x}{9y} \)

This is 0 when x = 0. Putting x = 0 in the equation of the curve:

    \( \displaystyle \frac{y^2}{16} = 1 \)

    \(y^2 = 16,\quad y = \pm 4\)

So the points are (0, 4) and (0, −4), the ends of the major axis of the ellipse.