MHT CET · Maths · Application of Derivatives
The equation of the tangent to the curve \(x=\operatorname{acos}^3 \theta, y=\operatorname{asin}^3 \theta\) at \(\theta=\frac{\pi}{4}\) is
- A \(x+y=\frac{\mathrm{a}}{\sqrt{2}}\)
- B \(x+y=\frac{\mathrm{a}}{2}\)
- C \(x+y=\frac{\mathrm{a}}{2 \sqrt{2}}\)
- D \(x+y=\frac{\mathrm{a}}{8}\)
Answer & Solution
Correct Answer
(A) \(x+y=\frac{\mathrm{a}}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\(x=\operatorname{acos}^3 \theta \text { and } y=\operatorname{asin}^3 \theta \)
\( \therefore \frac{\mathrm{~d} x}{\mathrm{~d} \theta}=-3 \operatorname{acos}^2 \theta \sin \theta \text { and } \frac{\mathrm{d} y}{\mathrm{~d} \theta}=3 \operatorname{asin}^2 \theta \cos \theta \)
\( \therefore \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{~d} \theta}}{\frac{\mathrm{~d} x}{\mathrm{~d} \theta}}=-\tan \theta \)
\( \therefore \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{\left(\theta=\frac{\pi}{4}\right)}=-1 \)
\( \text { At } \theta=\frac{\pi}{4} \)
\( x=\mathrm{a} \cos ^3 \frac{\pi}{4}=\frac{\mathrm{a}}{2 \sqrt{2}} \)
\( y =a \sin ^3 \frac{\pi}{4}=\frac{\mathrm{a}}{2 \sqrt{2}}\)
\(\therefore\) Equation of the tangent at \(\left(\frac{\mathrm{a}}{2 \sqrt{2}}, \frac{\mathrm{a}}{2 \sqrt{2}}\right)\) is
\(\begin{aligned}
& y-\frac{a}{2 \sqrt{2}}=-1\left(x-\frac{a}{2 \sqrt{2}}\right) \\
& \Rightarrow x+y=\frac{a}{\sqrt{2}}
\end{aligned}\)
\( \therefore \frac{\mathrm{~d} x}{\mathrm{~d} \theta}=-3 \operatorname{acos}^2 \theta \sin \theta \text { and } \frac{\mathrm{d} y}{\mathrm{~d} \theta}=3 \operatorname{asin}^2 \theta \cos \theta \)
\( \therefore \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{~d} \theta}}{\frac{\mathrm{~d} x}{\mathrm{~d} \theta}}=-\tan \theta \)
\( \therefore \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{\left(\theta=\frac{\pi}{4}\right)}=-1 \)
\( \text { At } \theta=\frac{\pi}{4} \)
\( x=\mathrm{a} \cos ^3 \frac{\pi}{4}=\frac{\mathrm{a}}{2 \sqrt{2}} \)
\( y =a \sin ^3 \frac{\pi}{4}=\frac{\mathrm{a}}{2 \sqrt{2}}\)
\(\therefore\) Equation of the tangent at \(\left(\frac{\mathrm{a}}{2 \sqrt{2}}, \frac{\mathrm{a}}{2 \sqrt{2}}\right)\) is
\(\begin{aligned}
& y-\frac{a}{2 \sqrt{2}}=-1\left(x-\frac{a}{2 \sqrt{2}}\right) \\
& \Rightarrow x+y=\frac{a}{\sqrt{2}}
\end{aligned}\)
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