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

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

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

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 $(4,4)$.

$$\begin{aligned} \text{We have,}\quad & y=4-x^2 \quad\text{.... (i)}\\ \text{and}\quad & y=x^2\quad\text{.... (ii)} \end{aligned}$$

$$\begin{array}{ll} \Rightarrow & \frac{d y}{d x}=-2 x \\ \text { and } & \frac{d y}{d x}=2 x \end{array}$$

$$\begin{array}{ll} \Rightarrow & m_1=-2 x \\ \text { and } & m_2=2 x \\ \text { From Eqs. (i) and (ii), } & x^2=4-x^2 \end{array}$$

$$\begin{array}{lr} \Rightarrow & 2 x^2=4 \\ \Rightarrow & x^2=2 \\ \Rightarrow & x= \pm \sqrt{2} \\ \therefore & y=x^2=( \pm \sqrt{2})^2=2 \end{array}$$

So, the points of intersection are $(\sqrt{2}, 2)$ and $(-\sqrt{2}, 2)$.

For point $(+\sqrt{2}, 2)$,

$$\begin{aligned} & m_1=-2 x=-2 \cdot \sqrt{2}=-2 \sqrt{2} \\ \text{and}\quad & m_2=2 x=2 \sqrt{2} \end{aligned}$$

and for point $(\sqrt{2}, 2), \quad \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{-2 \sqrt{2}-2 \sqrt{2}}{1-2 \sqrt{2} \cdot 2 \sqrt{2}}\right|=\left|\frac{-4 \sqrt{2}}{-7}\right|$

$$\therefore \quad \theta=\tan ^{-1}\left(\frac{4 \sqrt{2}}{7}\right)$$

We have, $y^2=4 x$ and $x^2+y^2-6 x+1=0$

Since, both the curves touch each other at (1, 2) i.e., curves are passing through $(1,2)$.

$$\begin{array}{lr} \therefore & 2 y \cdot \frac{d y}{d x}=4 \\ \text { and } & 2 x+2 y \frac{d y}{d x}=6 \end{array}$$

$$\begin{array}{ll} \Rightarrow & \frac{d y}{d x}=\frac{4}{2 y} \\ \text { and } & \frac{d y}{d x}=\frac{6-2 x}{2 y} \end{array}$$

$$\begin{array}{ll} \Rightarrow & \left(\frac{d y}{d x}\right)_{(1,2)}=\frac{4}{4}=1 \\ \text { and } & \left(\frac{d y}{d x}\right)_{(1,2)}=\frac{6-2 \cdot 1}{2 \cdot 2}=\frac{4}{4}=1 \end{array}$$

$\Rightarrow \quad m_1=1$ and $m_2=1$

Thus, we see that slope of both the curves are equal to each other i.e., $m_1=m_2=1$ at the point (1, 2).

Hence, both the curves touch each other.

Given equation of the curve is

$$3 x^2-y^2=8\quad\text{.... (i)}$$

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

$$\begin{aligned} 6 x-2 y \frac{d y}{d x} & =0 \\ \Rightarrow\quad \frac{d y}{d x}=\frac{6 x}{2 y} & =\frac{3 x}{y} \\ \Rightarrow\quad m_1 & =\frac{3 x}{y}\quad\text{[say]} \end{aligned}$$

and slope of normal $\left(m_2\right)=\frac{-1}{m_1}=\frac{-y}{3 x}\quad\text{.... (ii)}$

Since, slope of normal to the curve should be equal to the slope of line $x+3 y=4$, which is parallel to curve.

For line, $$y=\frac{4-x}{3}=\frac{-x}{3}+\frac{4}{3}$$

$\Rightarrow \quad$ Slope of the line $\left(m_3\right)=\frac{-1}{3}$

$$\begin{array}{lr} \therefore & m_2=m_3 \\ \Rightarrow & \frac{-y}{3 x}=-\frac{1}{3} \\ \Rightarrow & -3 y=-3 x \\ \Rightarrow & y=x\quad\text{.... (iii)} \end{array}$$

$$\begin{aligned} &\text { On substituting the value of } y \text { in Eq. (i), we get }\\ &\begin{array}{rlrl} & 3 x^2-x^2 =8 \\ \Rightarrow & x^2 =4 \\ \Rightarrow & x = \pm 2 \end{array} \end{aligned}$$

For $x=2$, $y=2\quad$ [using Eq. (iii)]

and for $x=-2$, $y=-2\quad$ [using Eq. (iii)]

Thus,the points at which normal to the curve are parallel to the line $x+3 y=4$ are $(2,2)$ and $(-2,-2)$.

Required equations of normal are

$$y-2=m_2(x-2) \text { and } y+2=m_2(x+2)$$

$$\begin{array}{llll} \Rightarrow & y-2=\frac{-2}{6}(x-2) \quad \text { and } y+2=\frac{-2}{6}(x+2) \\ \Rightarrow & 3 y-6=-x+2 \quad \text { and } 3 y+6=-x-2 \\ \Rightarrow & 3 y+x=+8 \quad \text { and } 3 y+x=-8 \end{array}$$

So, the required equations are $3 y+x= \pm 8$.

$$\begin{aligned} &\text { Given, equation of curve which is }\\ &\begin{aligned} & x^2+y^2-2 x-4 y+1=0 \quad\text{.... (i)}\\ \Rightarrow \quad & 2 x+2 y \frac{d y}{d x}-2-4 \frac{d y}{d x}=0 \end{aligned} \end{aligned}$$

$$\begin{aligned} \Rightarrow \quad & \frac{d y}{d x}(2 y-4) =2-2 x \\ \Rightarrow \quad & \frac{d y}{d x} =\frac{2(1-x)}{2(y-2)} \end{aligned}$$

$$\begin{aligned} &\text { Since, the tangents are parallel to the } Y \text {-axis i.e., } \tan \theta=\tan 90^{\circ}=\frac{d y}{d x} \text {. }\\ &\begin{array}{lr} \therefore & \frac{1-x}{y-2}=\frac{1}{0} \\ \Rightarrow & y-2=0 \\ \Rightarrow & y=2 \end{array} \end{aligned}$$

For $y=2$ from Eq. (i), we get

$$\begin{array}{rr} & x^2+2^2-2 x-4 \times 2+1=0 \\ \Rightarrow & x^2-2 x-3=0 \\ \Rightarrow & x^2-3 x+x-3=0 \\ \Rightarrow & x(x-3)+1(x-3)=0 \\ \Rightarrow & (x+1)(x-3)=0 \\ \therefore & x=-1, x=3 \end{array}$$

So, the required points are $(-1,2)$ and $(3,2)$.