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The point(s) on the curve x29+y264=1\frac{x^2}{9} + \frac{y^2}{64} = 1, at which the tangents are parallel to x-axis are

Solution

✅ Correct Option: 3

Differentiating implicitly: 2x9+2y⋅y′64=0\frac{2x}{9} + \frac{2y \cdot y'}{64} = 0, giving y′=−64x9yy' = -\frac{64x}{9y}. Tangent parallel to x-axis means y′=0y' = 0, so x=0x = 0. Substituting in the ellipse: y264=1⇒y=±8\frac{y^2}{64} = 1 \Rightarrow y = \pm 8. The points are (0,±8)(0, \pm 8).

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