The point(s) on the curve \(\displaystyle\frac{x^2}{9}+\frac{y^2}{64}=1\), at which the tangents are parallel to x-axis are
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \((0, \pm 8)\)
Differentiating implicitly with respect to x:
\( \displaystyle \frac{2x}{9}+\frac{2y}{64}\,\frac{dy}{dx} = 0 \)
\( \displaystyle \frac{dy}{dx} = -\frac{64x}{9y} \)
A tangent parallel to the x-axis has slope 0, so x = 0. Putting x = 0 in the equation of the curve:
\( \displaystyle \frac{y^2}{64} = 1 \)
\(y = \pm 8\)
So the points are \((0, \pm 8)\), the ends of the major axis.
Why the other options are incorrect
Option (1) \((0, \pm 3)\) — Incorrect. Not on the curve: \(0 + \frac{9}{64} \ne 1\).
Option (2) \((\pm 8, 0)\) — Incorrect. Not on the curve: \(\frac{64}{9} + 0 \ne 1\).
Option (4) \((\pm 3, 0)\) — Incorrect. These are on the curve, but there y = 0 and the tangents are parallel to the y-axis.