CUET · MATHS · PYQ PAPER 2025
The point on the curve \(\frac{x^2}{4}+\frac{y^2}{9}=1\) at which the tangent to the curve is parallel to the \(x\)-aris is
- A (0,3)
- B (2,0)
- C (-2,0)
- D (2,3)
Answer & Solution
Correct Answer
(A) (0,3)
Step-by-step Solution
Detailed explanation
\(\frac{d}{dx}\left(\frac{x^2}{4}+\frac{y^2}{9}\right) = \frac{d}{dx}(1)\) \(\frac{2x}{4} + \frac{2y}{9}\frac{dy}{dx} = 0\) \(\frac{dy}{dx} = -\frac{9x}{4y}\) \(\frac{dy}{dx} = 0 \implies -\frac{9x}{4y} = 0 \implies x = 0\)…
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