If $\mathbf{x=\sqrt{a^{\sin ^{-1} t}}, y=\sqrt{a^{\cos ^{-1} t}}}$.Show that $\mathbf{\frac{d y}{d x}=-\frac{y}{x}}$
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Ans: The given parametric equations are $x=\sqrt{a^{\sin ^{-1} t}}$ and $y=\sqrt{a^{\cos ^{-1} t}}$.
Now, $x=\sqrt{a^{\sin ^{-1} t}}$ and $y=\sqrt{a^{\cos ^{-1} t}}$
$\Rightarrow \mathrm{x}=\left(\mathrm{a}^{\sin ^{-1} \mathrm{t}}\right)$ and $\mathrm{y}=\left(\mathrm{a}^{\cos ^{-1} \mathrm{t}}\right)^{\frac{1}{2}}$
$\Rightarrow \mathrm{x}=\mathrm{a}^{\frac{1}{2} \sin ^{-1 \mathrm{t}}}$ and $\mathrm{y}=\mathrm{a}^{\frac{1}{2} \cos ^{-1} \mathrm{t}}$
Therefore, first consider $x=a^{\frac{1}{2} \sin ^{-1} t}$.
Take logarithms on both sides of the equation.
Then, we have
$\log x=\frac{1}{2} \sin ^{-1}$ tloga .
Then, differentiating both sides of the equation with respect to $t$ gives
$\begin{aligned}&\frac{1}{x} \times \frac{d x}{d t}=\frac{1}{2} \log a \times \frac{d}{d t}\left(\sin ^{-1} t\right) \\&\Rightarrow \frac{d x}{d t}=\frac{x}{2} \log a \times \frac{1}{\sqrt{1-t^{2}}} \\&\text { Therefore, } \frac{d x}{d t}=\frac{x \log a}{2 \sqrt{1-t^{2}}} . \ldots \ldots \text { (1) }\end{aligned}$
Again, consider the equation $y=a^{\frac{1}{2} \cos ^{-1} t}$.
Take logarithm both sides of the equation.
Then, we have
$\log y=\frac{1}{2} \cos ^{-1} \text { tloga }$
Differentiating both sides of the equation with respect to $\mathrm{t}$ gives $\frac{1}{y} \times \frac{d x}{d t}=\frac{1}{2} \log a \times \frac{d}{d t}\left(\cos ^{-1} t\right)$
$\begin{aligned}&\frac{1}{y} \times \frac{d x}{d t}=\frac{1}{2} \log a \times \frac{d}{d t}\left(\cos ^{-1} t\right) \\&\Rightarrow \frac{d x}{d t}=\frac{y \log a}{2} \times\left(\frac{1}{\sqrt{1-t^{2}}}\right)\end{aligned}$
Therefore, $\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{-\mathrm{yloga}}{2 \sqrt{1-\mathrm{t}^{2}}}$.
$\frac{dy}{dx} = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})} = \frac{(\frac{-y \log a}{2\sqrt{1- t^2}})}{(\frac{x \log }{2 \sqrt{1 - t^2}})} = \frac{y}{x}$
Hence, $\frac{dy}{dx} = \frac{y}{x}$