KCET · Maths · Hyperbola
If \( x=a \cos ^{3} \theta, y=a \sin ^{3} \theta \), then \( 1+\left(\frac{d y}{d x}\right)^{2} \) is
- A \( \tan \theta \)
- B \( \tan ^{2} \theta \)
- C \( \sec ^{2} \theta \)
- D \( 11 \)
Answer & Solution
Correct Answer
(C) \( \sec ^{2} \theta \)
Step-by-step Solution
Detailed explanation
Given that \(x=a \cos ^{3} \theta\) and \(y=a \sin ^{3} \theta\)
Now, \(\frac{d x}{d \theta}=a\left(3 \cos ^{2} \theta\right)(-\sin \theta)\)
and \(\frac{d y}{d \theta}=a\left(3 \sin ^{2} \theta\right)(\cos \theta)\)
Thus, \(\frac{d y}{d x}=\frac{d y}{d \theta} \times \frac{d \theta}{d x}\)
\(=\frac{3 a \sin ^{2} \theta \cos \theta}{-3 a \cos ^{2} \theta \sin \theta}=\frac{-\sin \theta}{\cos \theta}\)
\(\Rightarrow \frac{d y}{d x}=-\tan \theta \rightarrow(1)\)
So, \(1+\left(\frac{d y}{d x}\right)^{2}=1+\left(-\tan ^{2} \theta\right)\)
\(=1+\tan ^{2} \theta=\sec ^{2} \theta\)
Now, \(\frac{d x}{d \theta}=a\left(3 \cos ^{2} \theta\right)(-\sin \theta)\)
and \(\frac{d y}{d \theta}=a\left(3 \sin ^{2} \theta\right)(\cos \theta)\)
Thus, \(\frac{d y}{d x}=\frac{d y}{d \theta} \times \frac{d \theta}{d x}\)
\(=\frac{3 a \sin ^{2} \theta \cos \theta}{-3 a \cos ^{2} \theta \sin \theta}=\frac{-\sin \theta}{\cos \theta}\)
\(\Rightarrow \frac{d y}{d x}=-\tan \theta \rightarrow(1)\)
So, \(1+\left(\frac{d y}{d x}\right)^{2}=1+\left(-\tan ^{2} \theta\right)\)
\(=1+\tan ^{2} \theta=\sec ^{2} \theta\)
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