Q69 / 160 · TS EAMCET 2017 · Mathematics
If $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then $\frac{d^2 y}{d x^2}=$
- Correct answerA$-\frac{b^4}{a^2 y^3}$
- B$\frac{b^2}{a y^2}$
- C$\frac{-b^3}{a^2 y^3}$
- D$\frac{b^3}{a^2 y^2}$
Solution
We have,
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
$$
Let $x=a \cos \theta, y=b \sin \theta$
$$
\begin{aligned}
\therefore \quad \frac{d x}{d \theta} & =-a \sin \theta, \frac{d y}{d \theta}=b \cos \theta \\
\frac{d y}{d x} & =-\frac{b}{a} \cot \theta
\end{aligned}
$$
On differentiating w.r.t. $\theta$, we get
$$
\begin{array}{rlrl}
& \frac{d^2 y}{d x^2} & =-\frac{b}{a}\left(-\operatorname{cosec}^2 \theta\right) \frac{d \theta}{d x} \\
\Rightarrow \quad & \frac{d^2 y}{d x^2} & =\frac{b \operatorname{cosec}^2 \theta}{-a^2 \sin \theta} \\
\Rightarrow \quad & \frac{d^2 y}{d x^2} & =-\frac{b}{a^2 \sin ^3 \theta} & {\left[\because \sin \theta=\frac{y}{b}\right]} \\
\Rightarrow \quad & \frac{d^2 y}{d x} & =\frac{-b^4}{a^2 y^3}
\end{array}
$$
Hence, the correct answer is option: 'A'.