The points on the curve \(\displaystyle\frac{x^2}{9}+\frac{y^2}{16}=1\) at which the tangents are parallel to x-axis are:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \((0, \pm 4)\)
A tangent is parallel to the x-axis where its slope is zero, that is where \(\frac{dy}{dx} = 0\). Differentiating the equation implicitly:
\( \displaystyle \frac{2x}{9} + \frac{2y}{16}\,\frac{dy}{dx} = 0 \)
\( \displaystyle \frac{dy}{dx} = -\frac{16x}{9y} \)
This is zero when x = 0. Putting x = 0 in the equation of the curve:
\( \displaystyle \frac{y^2}{16} = 1 \)
\(y = \pm 4\)
So the tangents are parallel to the x-axis at both points (0, 4) and (0, −4), the top and bottom ends of the ellipse.
Why the other options are incorrect
Option (2) \((0, 4)\) — Incorrect. The tangent is horizontal there, but (0, −4) also qualifies; this option gives only one of the two points.
Option (3) \((0, -4)\) — Incorrect. It misses the second point (0, 4).
Option (4) \((0, 0)\) — Incorrect. (0, 0) is not on the curve: 0 + 0 ≠ 1.