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Mathematics Question on Derivatives of Functions in Parametric Forms

If x2a2+y2b2=1,\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 , then d2ydx2=\frac{d^2 y}{dx^2} =

A

−b4a2y3\frac{-b^4}{a^2 y^3}

B

b2ay2\frac{b^2}{ay^2}

C

−b3a2y3\frac{-b^3}{a^2 y^3}

D

b3a2y2\frac{b^3}{a^2 y^2}

Answer

−b4a2y3\frac{-b^4}{a^2 y^3}

Explanation

Solution

We have,
x2a2+y2b2=1\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1
Letx=acos⁡θ,y=bsin⁡θx=a \cos \theta, y=b \sin \theta
∴dxdθ=−asin⁡θ,dydθ=bcos⁡θ\therefore \frac{d x}{d \theta}=-a \sin \theta, \frac{d y}{d \theta}=b \cos \theta
dydx=−bacot⁡θ\frac{d y}{d x}=-\frac{b}{a} \cot \theta
On differentiating w.r.t. θ\theta, we get
d2ydx2=−ba(−cosec⁡2θ)dθdx\frac{d^{2} y}{d x^{2}}=-\frac{b}{a}\left(-\operatorname{cosec}^{2} \theta\right) \frac{d \theta}{d x}
⇒d2ydx2=bcosec⁡2θ−a2sin⁡θ\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{b \operatorname{cosec}^{2} \theta}{-a^{2} \sin \theta}
⇒d2ydx2=−ba2sin⁡3θ\Rightarrow \frac{d^{2} y}{d x^{2}}=-\frac{b}{a^{2} \sin ^{3} \theta}
⇒d2ydx=−b4a2y3[∵sin⁡θ=yb]\Rightarrow \frac{d^{2} y}{d x}=\frac{-b^{4}}{a^{2} y^{3}} [\because \sin \theta=\frac{y}{b}]