The equation of normal to the curve $\mathrm{y}=\sin \left(\frac{\pi x}{4}\right)$ at the point $(2,5)$ is
The equation of normal to the curve $\mathrm{y}=\sin \left(\frac{\pi x}{4}\right)$ at the point $(2,5)$ is
x+y=5
y=5
x=2
x+y=2
Solution
we have, $y=\sin \frac{\pi x}{4}$
$\begin{array}{l}
\frac{d y}{d x}=\frac{\pi}{4} \cdot \cos \frac{\pi x}{4} \\
\left(\frac{d y}{d x}\right)_{(2,5)}=\frac{\pi}{4} \cos \frac{2 \pi}{4}=0
\end{array}$
Since slope of tangent is zero, it is parallel to $\mathrm{X}$ axis. So normal is parallel to $\mathrm{Y}$ axis. Hence required equation of normal is $\mathrm{x}=2$