Let the side of a cube be $x$ unit.

$$\therefore \quad \text { Volume of cube }(V)=x^3$$

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

$$\frac{d V}{d t}=3 x^2 \frac{d x}{d t}=k\quad$$ [constant]

$\Rightarrow \quad \frac{d x}{d t}=\frac{k}{3 x^2}\quad\text{.... (i)}$

Also, surface area of cube, $$S=6 x^2$$

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

$$\begin{array}{ll} & \frac{d S}{d t}=12 x \cdot \frac{d x}{d t} \\ \Rightarrow & \frac{d S}{d t}=12 x \cdot \frac{k}{3 x^2} \quad\text{[using Eq. (i)]}\\ \Rightarrow & \frac{d S}{d t}=\frac{12 k}{3 x}=4\left(\frac{k}{x}\right) \\ \Rightarrow & \frac{d S}{d t} \propto \frac{1}{x} \end{array}$$

Hence, the surface area of the cube varies inversely as the length of the side.

Since, $x$ and $y$ are the sides of two squares such that $y=x-x^2$.

$\therefore \quad$ Area of the first square $\left(A_1\right)=x^2$

and area of the second square $\left(A_2\right)=y^2=\left(x-x^2\right)^2$

$$\therefore \quad \frac{d A_2}{d t}=\frac{d}{d t}\left(x-x^2\right)^2=2\left(x-x^2\right)\left(\frac{d x}{d t}-2 x \cdot \frac{d x}{d t}\right)$$

$=\frac{d x}{d t}(1-2 x) 2\left(x-x^2\right)$

and $$\frac{d A_1}{d t}=\frac{d}{d t} x^2=2x \cdot \frac{d x}{d t}$$

$\therefore\quad \frac{d A_2}{d A_1}=\frac{d A_2 / d t}{d A_1 / d t}=\frac{\frac{d x}{d t} \cdot(1-2 x)\left(2 x-2 x^2\right)}{2 x \cdot \frac{d x}{d t}}$

$$\begin{aligned} & =\frac{(1-2 x) 2 x(1-x)}{2 x} \\ & =(1-2 x)(1-x) \\ & =1-x-2 x+2 x^2 \\ & =2 x^2-3 x+1 \end{aligned}$$

$$\begin{array}{rlrl} & \text { Given, equation of curves are } & 2 x =y^2 \quad\text{.... (i)}\\ & \text { and } & 2 x y =k \quad\text{.... (ii)}\\ & \Rightarrow & y =\frac{k}{2 x}\quad\text{[from Eq. (ii)]} \end{array}$$

$$\begin{array}{lrl} \text { From Eq. (i), } & 2 x & =\left(\frac{k}{2 x}\right)^2 \\ \Rightarrow & 8 x^3 & =k^2 \\ \Rightarrow & x^3 & =\frac{1}{8} k^2 \\ \Rightarrow & x & =\frac{1}{2} k^{2 / 3} \\ \therefore & y & =\frac{k}{2 x}=\frac{k}{2 \cdot \frac{1}{2} k^{2 / 3}}=k^{1 / 3} \end{array}$$

Thus, we get point of intersection of curves which is $\left(\frac{1}{2} k^{2 / 3}, k^{1 / 3}\right)$.

From Eqs. (i) and (ii),

$$2=2 y \frac{d y}{d x}$$

and $$2\left[x \cdot \frac{d y}{d x}+y \cdot 1\right]=0$$

$$\begin{array}{ll} \Rightarrow & \frac{d y}{d x}=\frac{1}{y} \\ \text { and } & \left(\frac{d y}{d x}\right)=\frac{-2 y}{2 x}=-\frac{y}{x} \end{array}$$

$\Rightarrow\quad\left(\frac{d y}{d x}\right)_{\left(\frac{1}{2} k^{2 / 3}, k^{1 / 3}\right)}=\frac{1}{k^{1 / 3}}\quad \text { [say } m_1 \text { ] }$

$\text{and}\quad\left(\frac{d y}{d x}\right)_{\left(\frac{1}{2} k^{2 / 3}, k^{1 / 3}\right)}=\frac{-k^{1 / 3}}{\frac{1}{2} k^{2 / 3}}=-2 k^{-1 / 3}\quad \text { [say } m_2 \text { ] }$

Since, the curves intersect orthogonally.

$$\begin{array}{lrl} \text { i.e., } & m_1 \cdot m_2 & =-1 \\ \Rightarrow & \frac{1}{k^{1 / 3}} \cdot\left(-2 k^{-1 / 3}\right) & =-1 \end{array}$$

$$\begin{array}{lr} \Rightarrow & -2 k^{-2 / 3}=-1 \\ \Rightarrow & \frac{2}{k^{2 / 3}}=1 \\ \Rightarrow & k^{2 / 3}=2 \\ \therefore & k^2=8 \end{array}$$

which is the required condition.

$$\begin{array}{r} \text{Given equation of curves are}\quad x y=4 \quad\text{.... (i)}\\ \text{and}\quad x^2+y^2=8 \quad\text{.... (ii)}\\ \Rightarrow\quad x \cdot \frac{d y}{d x}+y=0 \end{array}$$

and $$2 x+2 y \frac{d y}{d x}=0$$

$$\begin{array}{ll} \Rightarrow & \frac{d y}{d x}=\frac{-y}{x} \\ \text { and } & \frac{d y}{d x}=\frac{-2 x}{2 y} \\ \Rightarrow & \frac{d y}{d x}=\frac{-y}{x}=m_1 \quad\text{[say]}\\ \text { and } & \frac{d y}{d x}=\frac{-x}{y}=m_2\quad\text{[say]} \end{array}$$

$$\begin{aligned} &\text { Since, both the curves should have same slope. }\\ &\begin{array}{ll} \therefore & \frac{-y}{x}=\frac{-x}{y} \Rightarrow-y^2=-x^2 \\ \Rightarrow & x^2=y^2\quad\text{.... (iii)} \end{array} \end{aligned}$$

$$\begin{aligned} &\text { Using the value of } x^2 \text { in Eq. (ii), we get }\\ &\begin{aligned} y^2+y^2 & =8 \\ \Rightarrow\quad y^2 & =4 \Rightarrow y= \pm 2 \end{aligned} \end{aligned}$$

For $y=2, x=\frac{4}{2}=2$ and for $y=-2, x=\frac{4}{-2}=-2$

Thus, the required points of intersection are $(2,2)$ and $(-2,-2)$.

For $(2,2)$, $$m_1=\frac{-y}{x}=\frac{-2}{2}=-1$$

$$\begin{array}{ll} \text { and } & m_2=\frac{-x}{y}=\frac{-2}{2}=-1 \\ \because & m_1=m_2 \end{array}$$

$$\begin{aligned} &\begin{aligned} \text { For }(-2,-2) \text {, }\quad & m_1=\frac{-y}{x}=\frac{-(-2)}{-2}=-1 \\ \text{and}\quad & m_2=\frac{-x}{y}=\frac{-(-2)}{-2}=-1 \end{aligned} \end{aligned}$$

Thus, for both the intersection points, we see that slope of both the curves are same. Hence, the curves touch each other.

$$\begin{array}{r} \text{We have,}\quad \sqrt{x}+\sqrt{y}=4 \quad\text{.... (i)}\\ \Rightarrow\quad x^{1 / 2}+y^{1 / 2}=4 \end{array}$$

$$\begin{aligned} \Rightarrow \quad \frac{1}{2} \cdot \frac{1}{x^{1 / 2}}+\frac{1}{2} \cdot \frac{1}{y^{1 / 2}} \cdot \frac{d y}{d x} & =0 \\ \therefore \quad \frac{d y}{d x} & =-\frac{1}{2} \cdot x^{-1 / 2} \quad 2 \cdot y^{1 / 2} \\ & =-\sqrt{\frac{y}{x}} \end{aligned}$$

$$\begin{aligned} &\text { Since, tangent is equally inclined to the axes. }\\ &\begin{array}{ll} \therefore & \frac{d y}{d x}= \pm 1 \\ \Rightarrow & -\sqrt{\frac{y}{x}}= \pm 1 \\ \Rightarrow & \frac{y}{x}=1 \Rightarrow y=x \end{array} \end{aligned}$$

From Eq. (i),

$$\begin{aligned} \sqrt{y}+\sqrt{y} & =4 \\ 2 \sqrt{y} & =4 \end{aligned}$$

$$\begin{array}{lr} \Rightarrow & 2 \sqrt{ } y=4 \\ \Rightarrow & 4 y=16 \end{array}$$

$$\therefore \quad y=4 \text { and } x=4$$

When $\mathrm{y}=4$, then $x=4$

So, the required coordinates are $(4,4)$.