The equation of a normal to the curve $x=4 \sec \theta$ and $y=4 \tan ^{2} \theta$ at $\theta=\frac{\pi}{4}$…
The equation of a normal to the curve $x=4 \sec \theta$ and $y=4 \tan ^{2} \theta$ at $\theta=\frac{\pi}{4}$ is
- $x+y \sqrt{2}=7 \sqrt{2}$
- $2 \sqrt{2} x+y=8 \sqrt{2}$
- $\sqrt{2} x+y=7 \sqrt{2}$
- $x+2 \sqrt{2} y=12 \sqrt{2}$
Solution
$\begin{array}{rll} & x=4 \sec \theta & \text { and } & y=4 \tan ^{2} \theta \\ \therefore & \frac{\mathrm{dx}}{\mathrm{d} \theta}=4 \sec \theta \tan \theta & \text { and } & \frac{\mathrm{dy}}{\mathrm{d} \theta}=8 \tan \theta \cdot \sec ^{2} \theta\end{array}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=$ slope of tangent $=\frac{8 \tan \theta \sec ^{2} \theta}{4 \sec \theta \tan \theta}=2 \sec \theta$
At $\theta=\frac{\pi}{4}, \quad \frac{\mathrm{dy}}{\mathrm{dx}}=2 \sqrt{2} \Rightarrow$ slope of normal $=-\frac{1}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}=\frac{-1}{2 \sqrt{2}}$
At $\theta=\frac{\pi}{4}, \mathrm{x}=4 \sqrt{2}$ and $\mathrm{y}=4$
Hence eq. of normal is
$\begin{aligned} &(y-4)=\frac{-1}{2 \sqrt{2}}(x-4 \sqrt{2}) \\ \therefore \quad & 2 \sqrt{2}(y-4)=-x+4 \sqrt{2} \Rightarrow x+2 \sqrt{2} y=12 \sqrt{2} \end{aligned}$
Asked in: MHT CET 2020 (13 Oct Shift 1)
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