The point on the curve \(\displaystyle\frac{x^2}{4} + \displaystyle\frac{y^2}{9} = 1\) 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)
The tangent is parallel to the x-axis where \(\frac{dy}{dx} = 0\). Differentiating implicitly:
\( \displaystyle \frac{2x}{4} + \frac{2y}{9}\frac{dy}{dx} = 0 \)
\( \displaystyle \frac{dy}{dx} = -\frac{9x}{4y} \)
So \(\frac{dy}{dx} = 0\) gives \(x = 0\). On the curve:
\( \displaystyle \frac{y^2}{9} = 1,\quad y = \pm 3 \)
The points are (0, 3) and (0, −3); only (0, 3) is among the options.
Why the other options are incorrect
Option (2) (2, 0) — Incorrect. Here y = 0, so \(\frac{dy}{dx}\) is undefined: the tangent is parallel to the y-axis.
Option (3) (−2, 0) — Incorrect. Again y = 0, and the tangent is parallel to the y-axis.
Option (4) (2, 3) — Incorrect. \(\frac{4}{4} + \frac{9}{9} = 2 \ne 1\), so the point is not on the curve.