The equation of normal to the curve $y^2=8x$at  origin is _____.

OR

The radius of circle is increasing at the uniform rate of 3 cm/sec. At the instant when the radius of the circle is 2 cm, its area increases at the rate of _____ cm2/s.

Given that the equation of normal to the curve $y^2=8x$ at (0, 0).
Differentiate the equation $y^2=8x$ w.r.t. x,
$\therefore 2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=8$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4}{\mathrm{y}}$
Which is slope of the tangent m,
Then recall that it’s normal should be $-\frac{1}{\mathrm{m}}$
Therefore, $-\frac{1}{\mathrm{m}}=-\frac{1}{{\displaystyle \frac{4}{\mathrm{y}}}}=-\frac{\mathrm{y}}{4}$
Slope at point (0, 0) will be $\mathrm{m}\text{'}(0,0) =-\frac{1}{\mathrm{m}}=-\frac{0}{4}=0$
Now, the equation of normal at the origin will be,
$\begin{array}{l}(\mathrm{y}-0)={\mathrm{m}}^{\mathrm{\prime }}(\mathrm{x}-0)\\ \therefore \mathrm{y}=0\left(\mathrm{x}\right)\\ \therefore \mathrm{y}=0\left(\mathrm{x}\right)\end{array}$
Which is the equation of normal at the origin.

OR

Let assume the radius of the circle as r.

From the given question the radius is increasing at the uniform rate 3 cm/s 
Hence,
$\frac{\mathrm{dr}}{\mathrm{dt}}=3\mathrm{cm}/\mathrm{s}$
Now, we know that area of the circle $\mathrm{A}={\mathrm{\pi r}}^2$
Differentiate the area w.r.t. t,
$\begin{array}{l}\therefore \frac{\mathrm{dA}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{\pi r}}^2\right)\\ \therefore \frac{\mathrm{dA}}{\mathrm{dt}}=\mathrm{\pi }\frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{r}}^2\right)\\ \therefore \frac{\mathrm{dA}}{\mathrm{dt}}=2{\mathrm{\pi r}}^{2-1}\frac{\mathrm{dr}}{\mathrm{dt}}\\ \therefore \frac{\mathrm{dA}}{\mathrm{dt}}=2\mathrm{\pi r}\frac{\mathrm{dr}}{\mathrm{dt}}\\ \therefore \frac{\mathrm{dA}}{\mathrm{dt}}=2\mathrm{\pi r}\left(3\right)\left(\because \frac{\mathrm{dr}}{\mathrm{dt}}=3\right)\\ \therefore \frac{\mathrm{dA}}{\mathrm{dt}}=6\mathrm{\pi r}\end{array}$
Substitute r=2 cm
$\begin{array}{l}\therefore \frac{\mathrm{dA}}{\mathrm{dt}}=6\mathrm{\pi }\left(2\right)\\ \therefore \frac{\mathrm{dA}}{\mathrm{dt}}=12\mathrm{\pi }{\mathrm{cm}}^2/\mathrm{s}\end{array}$
Hence, area of the circle is increasing at the rate of $12\mathrm{\pi }{\mathrm{cm}}^2/\mathrm{s}$, when radius is 2 cm.