AP EAMCET · Maths · Application of Derivatives
The equation of tangent of the curve \(y=\sqrt{9-2 x^2}\) at the point where the ordinate and abscissa are equal is
- A \(2 x+y-3 \sqrt{3}=0\)
- B \(2 x+y+3 \sqrt{3}=0\)
- C \(2 x-y-3 \sqrt{3}=0\)
- D \(2 x-y+3 \sqrt{3}=0\)
Answer & Solution
Correct Answer
(A) \(2 x+y-3 \sqrt{3}=0\)
Step-by-step Solution
Detailed explanation
We have curve \(y=\sqrt{9-2 x^2}\) Let \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) be any point such that \(\mathrm{x}_1=\mathrm{y}_1\) on the curve \(\Rightarrow \mathrm{x}_1{ }^2=\mathrm{g}-2 \mathrm{x}_1{ }^2\)…
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