The ponts(s) on the curve\(\frac{x^2}{9}\ + \frac{y^2}{64}\) = 1, at which the tangents are parallel to x-axis are

(0,±8)

equation of curve = \(\frac{x^2}{9}\ + \frac{y^2}{64}\) = 1

On differentiating both sides with respect to x,

$\frac{2x}{9}+\frac{2y}{24}\frac{dy}{dx}-0⇒\frac{dy}{dx}=\frac{-64x}{9y}⇒as\, \frac{-\frac{2y}{64}}{\frac{2x}{9}}=\frac{-64x}{9y}$

So the tangent parallel to x-axis if slope of the tangent is 0.

i.e., $\frac{-64x}{9y}=0$, which is possible if x=0.

\(\frac{x^2}{9}\ + \frac{y^2}{64}\) = 1

$x = 0 ⇒ y^2 = 64 ⇒ ±8 = y$

points of tangent are (0, ±8)

So option 3 is correct.