Given, radius of the sphere is R.
Let r and h be the radius and the height of the inscribed cylinder respectively
We have:
h = $2\sqrt{{\mathit{R}}^2-{\mathit{r}}^2}$
Let Volume of cylinder = V
V = $\mathit{\pi }{\mathit{r}}^2\mathit{h}$

= $\mathit{\pi }{\mathit{r}}^2\times 2\sqrt{{\mathit{R}}^2-{\mathit{r}}^2}$
= $2\mathit{\pi }{\mathit{r}}^2\sqrt{{\mathit{R}}^2-{\mathit{r}}^2}$
Differentiating the above function w.r.t. r, we have,
V = $2\mathit{\pi }{\mathit{r}}^2\sqrt{{\mathit{R}}^2-{\mathit{r}}^2}$
$\frac{\mathit{d}\mathit{V}}{\mathit{d}\mathit{r}}$ = $4\mathit{\pi }\mathit{r}\sqrt{{\mathit{R}}^2-{\mathit{r}}^2}$ - $\frac{4\mathit{\pi }{\mathit{r}}^2}{2\sqrt{{\mathit{R}}^2-{\mathit{r}}^2}}$
= $\frac{4\mathit{\pi }\mathit{r}\left({\mathit{R}}^2-{\mathit{r}}^2-4\mathit{\pi }{\mathit{r}}^3\right)}{2\sqrt{{\mathit{R}}^2-{\mathit{r}}^2}}$
$\frac{\mathit{d}\mathit{V}}{\mathit{d}\mathit{r}}$ = $\frac{4\mathit{\pi }\mathit{r}{\mathit{R}}^2-4\mathit{\pi }{\mathit{r}}^3-2\mathit{\pi }{\mathit{r}}^3}{2\sqrt{{\mathit{R}}^2-{\mathit{r}}^2}}$
= $\frac{4\mathit{\pi }\mathit{r}{\mathit{R}}^2-6\mathit{\pi }{\mathit{r}}^3}{2\sqrt{{\mathit{R}}^2-{\mathit{r}}^2}}$
For maxima or minima, $\frac{\mathit{d}\mathit{V}}{\mathit{d}\mathit{r}}$ = 0 ⇒ $4\mathit{\pi }\mathit{r}{\mathit{R}}^2-6\mathit{\pi }{\mathit{r}}^3$ = 0
⇒ $6\mathit{\pi }{\mathit{r}}^3$ = $4\mathit{\pi }\mathit{r}{\mathit{R}}^2$
⇒ ${\mathit{r}}^2$ = $\frac{2{\mathit{R}}^2}{3}$
$\frac{\mathit{d}\mathit{V}}{\mathit{d}\mathit{r}}$ = $\frac{4\mathit{\pi }\mathit{r}{\mathit{R}}^2-6\mathit{\pi }{\mathit{r}}^3}{2\sqrt{{\mathit{R}}^2-{\mathit{r}}^2}}$
Now, $\frac{{\mathit{d}}^2\mathit{V}}{\mathit{d}{\mathit{r}}^2}$ =
=
=
Now, when ${\mathit{r}}^2$ = $\frac{2{\mathit{R}}^2}{3},\frac{{\mathit{d}}^2\mathit{V}}{\mathit{d}{\mathit{r}}^2}$ < 0
∴ Volume is the maximum when ${\mathit{r}}^2$ = $\frac{2{\mathit{R}}^2}{3}$
when ${\mathit{r}}^2$ = $\frac{2{\mathit{R}}^2}{3}$ , h = $2\sqrt{{\mathit{R}}^2-\frac{2{\mathit{R}}^2}{3}}$ = 2 $\sqrt{\frac{{\mathit{R}}^2}{3}}$ = $\frac{2\mathit{R}}{\sqrt{3}}$
Hence, the volume of the cylinder is the maximum when the height of the cylinder is $\frac{2\mathit{R}}{\sqrt{3}}$
OR
The equation of the given curve is ${\mathit{x}}^2$ = 4y.
Differentiating w.r.t. x, we get
$\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ = $\frac{\mathit{x}}{2}$
Let (h, k) be the co- ordinates of the point of contact of the normal to the curve ${\mathit{x}}^2$ = 4y.
Now, slope of the tangent at (h, k) is given by
$\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}{|}_{\left(\mathit{h},\mathit{k}\right)}$ = $\frac{\mathit{h}}{2}$
Hence, slope of the normal at (h , k) = $\frac{-2}{\mathit{h}}$
Therefore, the equation of normal at (h, k) is
y - k = $\frac{-2}{\mathit{h}}\left(\mathit{x}-\mathit{h}\right)$ ... (1)
Since, it passes through the point (1, 2) we have
2 - k = $\frac{-2}{\mathit{h}}\left(1-\mathit{h}\right)$ or k = 2 + $\frac{2}{\mathit{h}}$ (1 - h) ... (2)
Now, (h, k) lies on the curve ${\mathit{x}}^2$ = 4y, so, we have:
${\mathit{h}}^2$ = 4k ... (3)
Solving (2) and (3), we get,
h = 2 and k = 1.
From (1), the required equation of the normal is:
y - 1 = $\frac{-2}{2}$ (x - 2) or x + y = 3
Also, slope of the tangent = 1
∴ Equation of tangent at (1, 2) is:
y – 2 = 1(x – 1) or y = x + 1