$$\begin{array}{lr} \text { Given, } & \text { line is } x \cos \alpha+y \sin \alpha=p \quad\text{... (i)}\\ \text { and } & \text { curve is } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ \Rightarrow & b^2 x^2+a^2 y^2=a^2 b^2\quad\text{.... (ii)} \end{array}$$

$$\begin{aligned} &\text { Now, differentiating Eq. (ii) w.r.t. } x \text {, we get }\\ &b^2 \cdot 2 x+a^2 \cdot 2 y \cdot \frac{d y}{d x}=0 \end{aligned}$$

$$\begin{array}{lr} \Rightarrow & \frac{d y}{d x}=\frac{-2 b^2 x}{2 a^2 y}=\frac{-x b^2}{y a^2} \quad\text{.... (iii)}\\ \text { From Eq. (i), } & y \sin \alpha=p-x \cos \alpha \\ \Rightarrow & y=-x \cot \alpha+\frac{p}{\sin \alpha} \end{array}$$

Thus, slope of the line is $(-\cot \alpha)$.

So, the given equation of line will be tangent to the Eq. (ii), if $\left(-\frac{x}{y} \cdot \frac{b^2}{a^2}\right)=(-\cot \alpha)$

$$\begin{aligned} \Rightarrow \quad \frac{x}{a^2 \cos \alpha} & =\frac{y}{b^2 \sin \alpha}=k \quad\text{[say]}\\ \Rightarrow \quad x & =k a^2 \cos \alpha \\ \text { and } \quad y & =b^2 k \sin \alpha \end{aligned}$$

So, the line $x \cos \alpha+y \sin \alpha=p$ will touch the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}$ at point $\left(k a^2 \cos \alpha, k b^2 \sin \alpha\right)$

$\begin{aligned} & \text { From Eq. (i), } \quad k a^2 \cos ^2 \alpha+k b^2 \sin ^2 \alpha=p \\ & \Rightarrow \quad a^2 \cos ^2 \alpha+b^2 \sin ^2 \alpha=\frac{p}{k}\end{aligned}$

$\Rightarrow \quad\left(a^2 \cos ^2 \alpha+b^2 \sin ^2 \alpha\right)^2=\frac{p^2}{k^2}\quad\text{.... (iv)}$

$$\begin{array}{lrl} \text { From Eq. (ii), } & b^2 k^2 a^4 \cos ^2 \alpha+a^2 k^2 b^4 \sin ^2 \alpha & =a^2 b^2 \\ \Rightarrow & k^2\left(a^2 \cos ^2 \alpha+b^2 \sin ^2 \alpha\right) & =1 \\ \Rightarrow & \left(a^2 \cos ^2 \alpha+b^2 \sin ^2 \alpha\right) & =\frac{1}{k^2}\quad\text{.... (v)} \end{array}$$

$$\begin{aligned} &\text { On dividing Eq. (iv) by Eq. (v), we get }\\ &a^2 \cos ^2 \alpha+b^2 \sin ^2 \alpha=p^2 \end{aligned}$$

Hence proved.

Alternate Method

We know that, if a line $y=m x+c$ touches ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then the required condition is $c^2=a^2 m^2+b^2$

Here, given equation of the line is

$x \cos \alpha+y \sin \alpha=p$

$$ \begin{aligned} \Rightarrow \quad &y =\frac{p-x \cos \alpha}{\sin \alpha} \\ & =-x \cot \alpha+\frac{p}{\sin \alpha} \end{aligned}$$

$$\begin{array}{ll} \Rightarrow & c=\frac{p}{\sin \alpha} \\ \text { and } & m=-\cot \alpha \end{array}$$

$\therefore\quad\left(\frac{p}{\sin \alpha}\right)^2=a^2(-\cot \alpha)^2+b^2$

$$\begin{array}{ll} \Rightarrow & \frac{p^2}{\sin ^2 \alpha}=a^2 \frac{\cos ^2 \alpha}{\sin ^2 \alpha}+b^2 \\ \Rightarrow & p^2=a^2 \cos ^2 \alpha+b^2 \sin ^2 \alpha\quad\text{Hence proved.} \end{array}$$

Let the length of side of the square base of open box be $x$ units and its height be $y$ units.

$$\begin{array}{lr} \therefore & \text { Area of the metal used }=x^2+4 x y \\ \Rightarrow & x^2+4 x y=c^2 \quad\text{[given]}\\ \Rightarrow & y=\frac{c^2-x^2}{4 x}\quad\text{.... (i)} \end{array}$$

$$\begin{aligned} &\quad \text { Now, } \quad \text { volume of the box }(V)=x^2 y\\ &\begin{aligned} \Rightarrow \quad V & =x^2 \cdot\left(\frac{c^2-x^2}{4 x}\right) \\ & =\frac{1}{4} x\left(c^2-x^2\right) \\ & =\frac{1}{4}\left(c^2 x-x^3\right) \end{aligned} \end{aligned}$$

On differentiating both sides w.r.t. $x$, we get

$$\begin{aligned} & \frac{d V}{d x}=\frac{1}{4}\left(c^2-3 x^2\right) \\ \text{Now,}\quad & \frac{d V}{d x}=0 \Rightarrow c^2=3 x^2 \end{aligned}$$

$$ \begin{aligned} &\begin{array}{ll} \Rightarrow & x^2=\frac{c^2}{3} \\ \Rightarrow & x=\frac{c}{\sqrt{3}}\quad\text{[using positive sign]} \end{array}\\ &\text { Again, differentiating Eq. (ii) w.r.t. } x \text {, we get } \end{aligned}$$

$$\begin{aligned} \frac{d^2 V}{d x^2} & =\frac{1}{4}(-6 x)=\frac{-3}{2} x<0 \\ \therefore \quad\left(\frac{d^2 v}{d x^2}\right)_{\text {at } x=\frac{c}{\sqrt{3}}} & =-\frac{3}{2} \cdot\left(\frac{c}{\sqrt{3}}\right)<0 \end{aligned}$$

Thus, we see that volume $(V)$ is maximum at $x=\frac{c}{\sqrt{3}}$.

$\therefore$ Maximum volume of the box, $(V)_{x=\frac{c}{\sqrt{3}}}=\frac{1}{4}\left(c^2 \cdot \frac{c}{\sqrt{3}}-\frac{c^3}{3 \sqrt{3}}\right)$

$$\begin{aligned} & =\frac{1}{4} \cdot \frac{\left(3 c^3-c^3\right)}{3 \sqrt{3}}=\frac{1}{4} \cdot \frac{2 c^3}{3 \sqrt{3}} \\ & =\frac{c^3}{6 \sqrt{3}} \text { cu units } \end{aligned}$$

Let breadth and length of the rectangle be $x$ and $y$, respectively.

$$\begin{array}{lr} \because & \text { Perimeter of the rectangle }=36 \mathrm{~cm} \\ \Rightarrow & 2 x+2 y=36 \\ \Rightarrow & x+y=18 \\ \Rightarrow & y=18-x \quad\text{.... (i)} \end{array}$$

Let the rectangle is being revolved about its length $y$.

Then, volume $(V)$ of resultant cylinder $=\pi x^2 \cdot y$

$\Rightarrow \quad V=\pi x^2 \cdot(18-x) \quad\left[\because V=\pi r^2 h\right][$ [using Eq. (i) $]$

$=18 \pi x^2-\pi x^3=\pi\left[18 x^2-x^3\right]$

On differentiating both sides w.r.t. $x$, we get

$$\begin{array}{llrl} & \frac{d V}{d x} =\pi\left(36 x-3 x^2\right) \\ \text { Now, } & \frac{d V}{d x} =0 \\ \Rightarrow & 36 x =3 x^2 \end{array}$$

$$\begin{aligned} \Rightarrow \quad& 3 x^2-36 x =0 \\ \Rightarrow \quad& 3\left(x^2-12 x\right) =0 \\ \Rightarrow \quad& 3 x(x-12) =0 \\ \Rightarrow \quad& x =0, x=12 \\ \therefore \quad& x =12\quad [\because, x \neq 0] \end{aligned}$$

$$\begin{aligned} &\text { Again, differentiating w.r.t. } x \text {, we get }\\ &\begin{aligned} \frac{d^2 V}{d x^2} & =\pi(36-6 x) \\ \Rightarrow \quad\left(\frac{d^2 V}{d x^2}\right)_{x=12} & =\pi(36-6 \times 12)=-36 \pi<0 \end{aligned} \end{aligned}$$

At $x=12$, volume of the resultant cylinder is the maximum.

So, the dimensions of rectangle are 12 cm and 6 cm, respectively. $\quad$ [using Eq. (i)]

$$\begin{aligned} &\therefore \text { Maximum volume of resultant cylinder, }\\ &\begin{aligned} (V)_{x=12} & =\pi\left[18 \cdot(12)^2-(12)^3\right] \\ & =\pi\left[12^2(18-12)\right] \\ & =\pi \times 144 \times 6 \\ & =864 \pi \mathrm{~cm}^3 \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { Let length of one edge of cube be } x \text { units and radius of sphere be } r \text { units. }\\ &\begin{aligned} & \therefore \quad \text { Surface area of cube }=6 x^2 \\ & \text { and } \quad \text { surface area of sphere }=4 \pi r^2 \end{aligned} \end{aligned}$$

Also, $$6 x^2+4 \pi r^2=k\quad$$ [constant, given]

$$\begin{array}{ll} \Rightarrow & 6 x^2=k-4 \pi r^2 \\ \Rightarrow & x^2=\frac{k-4 \pi r^2}{6} \\ \Rightarrow & x=\left[\frac{k-4 \pi r^2}{6}\right]^{1 / 2}\quad\text{.... (i)} \end{array}$$

Now, volume of cube $=x^3$

and volume of sphere $=\frac{4}{3} \pi r^3$

Let sum of volume of the cube and volume of the sphere be given by

$$S=x^3+\frac{4}{3} \pi r^3=\left[\frac{k-4 \pi r^2}{6}\right]^{3 / 2}+\frac{4}{3} \pi r^3$$

On differentiating both sides w.r.t. $r$, we get

$$\begin{aligned} \frac{d S}{d r} & =\frac{3}{2}\left[\frac{k-4 \pi r^2}{6}\right]^{1 / 2} \cdot\left(\frac{-8 \pi r}{6}\right)+\frac{12}{3} \pi r^2 \\ & =-2 \pi r\left[\frac{k-4 \pi r^2}{6}\right]^{1 / 2}+4 \pi r^2 \quad\text{.... (ii)}\\ & =-2 \pi r\left[\left\{\frac{k-4 \pi r^2}{6}\right\}^{1 / 2}-2 r\right] \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text { Now, }\quad\frac{d S}{d r} & =0 \\ \Rightarrow\quad r & =0 \text { or } 2 r=\left(\frac{k-4 \pi r^2}{6}\right)^{1 / 2} \end{aligned} \end{aligned}$$

$$\begin{array}{ll} \Rightarrow & 4 r^2=\frac{k-4 \pi r^2}{6} \Rightarrow 24 r^2=k-4 \pi r^2 \\ \Rightarrow & 24 r^2+4 \pi r^2=k \Rightarrow r^2[24+4 \pi]=k \\ \therefore & r=0 \text { or } \quad r=\sqrt{\frac{k}{24+4 \pi}}=\frac{1}{2} \sqrt{\frac{k}{6+\pi}} \end{array}$$

$\therefore \quad r=0$ or $r=\sqrt{\frac{k}{24+4 \pi}}=\frac{1}{2} \sqrt{\frac{k}{6+\pi}}$

$$\begin{aligned} &\begin{aligned} \text { We know that, }\quad & r \neq 0 \\ \therefore\quad & r=\frac{1}{2} \sqrt{\frac{k}{6+\pi}} \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { Again, differentiating w.r.t. } r \text { in Eq. (ii), we get }\\ &\frac{d^2 S}{d r^2}=\frac{d}{d r}\left[-2 \pi\left\{\left(\frac{k-4 \pi r^2}{6}\right)^{1 / 2}+4 \pi r^2\right\}\right] \end{aligned}$$

$$\begin{aligned} & =-2 \pi\left[r \cdot \frac{1}{2}\left(\frac{k-4 \pi r^2}{6}\right)^{-1 / 2} \cdot\left(\frac{-8 \pi r}{6}\right)+\left(\frac{k-4 \pi r^2}{6}\right)^{1 / 2} \cdot 1\right]+4 \pi \cdot 2 r \\ & =-2 \pi\left[r \cdot \frac{1}{2 \sqrt{\frac{k-4 \pi r^2}{6}}} \cdot\left(\frac{-8 \pi r}{6}\right)+\sqrt{\frac{k-4 \pi r^2}{6}}\right]+8 \pi r \end{aligned}$$

$$\begin{aligned} & =-2 \pi\left[\frac{-8 \pi r^2+12\left(k-\frac{4 \pi r^2}{6}\right)}{12 \sqrt{\frac{k-4 \pi r^2}{6}}}\right]+8 \pi r \\ & =-2 \pi\left[\frac{-48 \pi r^2+72 k-48 \pi r^2}{72 \sqrt{\frac{k-4 \pi r^2}{6}}}\right]+8 \pi r=-2 \pi\left[\frac{-96 \pi r^2+72 k}{72 \sqrt{\frac{k-4 \pi r^2}{6}}}\right]+8 \pi r>0 \end{aligned}$$

For $r=\frac{1}{2} \sqrt{\frac{k}{6+\pi}}$, then the sum of their volume is minimum.

For $r=\frac{1}{2} \sqrt{\frac{k}{6+\pi}}\quad$, $$x=\left[\frac{k-4 \pi \cdot \frac{1}{4} \frac{k}{(6+\pi)}}{6}\right]^{1 / 2}$$

$=\left[\frac{(6+\pi) k-\pi k}{6(6+\pi)}\right]^{1 / 2}=\left[\frac{k}{6+\pi}\right]^{1 / 2}=2 r$

Since, the sum of their volume is minimum when $x=2 r$.

Hence, the ratio of an edge of cube to the diameter of the sphere is $1: 1$.

We have, $$A B=2 r$$

and $\angle A C B=90^{\circ} \quad$ [since, angle in the semi-circle is always $90^{\circ}$ ]

Let $A C=x$ and $B C=y$

$$\begin{array}{ll} \therefore & (2 r)^2=x^2+y^2 \\ \Rightarrow & y^2=4 r^2-x^2 \\ \Rightarrow & y=\sqrt{4 r^2-x^2}\quad\text{.... (i)} \end{array}$$

$$\begin{aligned} \text { Now, } \quad \quad \text { area of } \triangle A B C, A & =\frac{1}{2} \times x \times y \\ & =\frac{1}{2} \times x \times\left(4 r^2-x^2\right)^{1 / 2}\quad\text{[using Eq. (i)]} \end{aligned}$$

$$\begin{aligned} &\text { Now, differentiating both sides w.r.t. } x \text {, we get }\\ &\begin{aligned} \frac{d A}{d x} & =\frac{1}{2}\left[x \cdot \frac{1}{2}\left(4 r^2-x^2\right)^{-1 / 2} \cdot(0-2 x)+\left(4 r^2-x^2\right)^{1 / 2} \cdot 1\right] \\ & =\frac{1}{2}\left[\frac{-2 x^2}{2 \sqrt{4 r^2-x^2}}+\left(4 r^2-x^2\right)^{1 / 2}\right] \end{aligned} \end{aligned}$$

$$\begin{aligned} & =\frac{1}{2}\left[\frac{-x^2}{\sqrt{4 r^2-x^2}}+\sqrt{4 r^2-x^2}\right] \\ & =\frac{1}{2}\left[\frac{-x^2+4 r^2-x^2}{\sqrt{4 r^2-x^2}}\right]=\frac{1}{2}\left[\frac{-2 x^2+4 r^2}{\sqrt{4 r^2-x^2}}\right] \end{aligned}$$

$$\begin{array}{lrl} \Rightarrow & \frac{d A}{d x} & =\left[\frac{\left(-x^2+2 r^2\right)}{\sqrt{4 r^2-x^2}}\right] \\ \text { Now, } & \frac{d A}{d x} & =0 \\ \Rightarrow & -x^2+2 r^2 & =0 \\ \Rightarrow & r^2 & =\frac{1}{2} x^2 \\ \Rightarrow & r & =\frac{1}{\sqrt{2}} x \\ \therefore & x & =r \sqrt{2} \end{array}$$

Again, differentiating both sides w.r.t. x, we get

$$\begin{aligned} \frac{d^2 A}{d x^2} & =\frac{\sqrt{4 r^2-x^2} \cdot(-2 x)+\left(2 r^2-x^2\right) \cdot \frac{1}{2}\left(4 r^2-x^2\right)^{-1 / 2}(-2 x)}{\left(\sqrt{4 r^2-x^2}\right)^2} \\ & =\frac{-2 x\left[\sqrt{4 r^2-x^2}+\left(2 r^2-x^2\right) \cdot \frac{1}{2 \sqrt{4 r^2-x^2}}\right]}{\left(\sqrt{4 r^2-x^2}\right)^2} \\ & =\frac{-4 x \cdot\left(\sqrt{4 r^2-x^2}\right)^2+\left(2 r^2-x^2\right)(-2 x)}{2 \cdot\left(4 r^2-x^2\right)^{3 / 2}} \\ & =\frac{-4 x\left(4 r^2-x^2\right)+\left(2 r^2-x^2\right) \cdot(-2 x)}{2 \cdot\left(4 r^2-x^2\right)^{3 / 2}} \\ & =\frac{-16 x r^2+4 x^3+\left(2 r^2-x^2\right)(-2 x)}{2 \cdot\left(4 r^2-x^2\right)^{3 / 2}} \end{aligned}$$

$$\begin{aligned} \left(\frac{d^2 A}{d x^2}\right)_{x=r \sqrt{2}} & =\frac{-16 \cdot r \sqrt{2} \cdot r^2+4 \cdot(r \sqrt{2})^3+\left[2 r^2-(r \sqrt{2})^2\right] \cdot(-2 \cdot r \sqrt{2})}{2 \cdot\left(4 r^2-2 r^2\right)^{3 / 2}} \quad[\because x=r \sqrt{2}] \\ & =\frac{-16 \sqrt{2} \cdot r^3+8 \sqrt{2} r^3}{2\left(2 r^2\right)^{3 / 2}}=\frac{8 \sqrt{2} r^2[r-2 r]}{4 r^3} \\ & =\frac{-8 \sqrt{2} r^3}{4 r^3}=-2 \sqrt{2}<0 \end{aligned}$$

For $x=r \sqrt{2}$, the area of triangle is maximum.

For $x=r \sqrt{2}\quad, y=\sqrt{4 r^2-(r \sqrt{2})^2}=\sqrt{2 r^2}=r \sqrt{2}$

Since, $\quad x=r \sqrt{2}=y$

Hence, the triangle is isosceles.