A) \[\frac{a}{{{b}^{2}}}{{\sec }^{2}}\theta \]
B) \[\frac{-b}{a}{{\sec }^{2}}\theta \]
C) \[\frac{-b}{{{a}^{2}}}{{\sec }^{3}}\theta \]
D) \[\frac{-b}{{{a}^{2}}}{{\sec }^{3}}\theta \]
Correct Answer: C
Solution :
\[\frac{dx}{d\theta }=a\cos \theta \] and \[\frac{dy}{d\theta }=-b\sin \theta \] Þ \[\frac{dy}{dx}=\frac{-b}{a}\tan \theta \] and \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{-b}{a}{{\sec }^{2}}\theta \frac{d\theta }{dx}\] Þ \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{-b}{a}{{\sec }^{2}}\theta \frac{1}{a\cos \theta }=\frac{-b}{{{a}^{2}}}{{\sec }^{3}}\theta \].
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