MHT CET · Maths · Application of Derivatives
The equation of tangent to the curve given by \(x=3 \cos \theta, y=3 \sin \theta\) at \(\theta=\frac{\pi}{4}\) is
- A \(x+y=\sqrt{2}\)
- B \(3 x+y=3 \sqrt{2}\)
- C \(x+y=3 \sqrt{2}\)
- D \(x+3 y=3 \sqrt{2}\)
Answer & Solution
Correct Answer
(C) \(x+y=3 \sqrt{2}\)
Step-by-step Solution
Detailed explanation
Given, \(x=3 \cos \theta, y=3 \sin \theta\). On squaring and
adding we get \(x^{2}+y^{2}=9\), which represent a circle.
Equation of tangent at \(\theta=\frac{\pi}{4}\) is
\(x \cdot\left(3 \cos \frac{\pi}{4}\right)+y \cdot\left(3 \sin \frac{\pi}{4}\right)=9 \)
\( \Rightarrow \quad x+y=3 \sqrt{2} \)
adding we get \(x^{2}+y^{2}=9\), which represent a circle.
Equation of tangent at \(\theta=\frac{\pi}{4}\) is
\(x \cdot\left(3 \cos \frac{\pi}{4}\right)+y \cdot\left(3 \sin \frac{\pi}{4}\right)=9 \)
\( \Rightarrow \quad x+y=3 \sqrt{2} \)
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