(i) Given that, directrix $=0$ and focus $=(6,0)$

So, the equation of the parabola

$$\begin{array}{rlrl} & (x-6)^2+y^2 =x^2 \\ \Rightarrow & x^2+36-12 x+y^2 =x^2 \\ \Rightarrow & y^2-12 x+36 =0 \end{array}$$

(ii) Given that, vertex $=(0,4)$ and focus $=(0,2)$

So, the equation of parabola is

$$\begin{array}{rlrl} & \sqrt{(x-0)^2+(y-2)^2} =|y-6| \\ \Rightarrow & x^2+y^2-4 y+4 =y^2-12 y+36 \\ \Rightarrow & x^2-4 y+12 y-32 =0 \\ \Rightarrow & x^2+8 y-32 =0 \\ \Rightarrow & x^2 =32-8 y \end{array}$$

(iii) Given that, focus at $(-1,-2)$ and directrix $x-2 y+3=0$

So, the equation of parabola is $\sqrt{(x+1)^2+(y+2)^2}=\left|\frac{x-2 y+3}{\sqrt{1+4}}\right|$

$$\begin{array}{lrl} \Rightarrow & x^2+2 x+1+y^2+4 y+4 & =\frac{1}{5}\left[x^2+4 y^2+9-4 x y-12 y+6 x\right] \\ \Rightarrow & 4 x^2+4 x y+y^2+4 x+32 y+16 & =0 \end{array}$$

Let the coordinates of the point be $(x, y)$, then according to the question,

$$\begin{array}{ll} & \sqrt{(x-3)^2+y^2}+\sqrt{(x-9)^2+y^2}=12 \\ \Rightarrow & \sqrt{(x-3)^2+y^2}=12-\sqrt{(x-9)^2+y^2} \end{array}$$

$$\begin{aligned} &\text { On squaring both sides, we get }\\ &\begin{array}{llrl} & x^2-6 x+9+y^2 =144+\left(x^2-18 x+81+y^2\right)-24 \sqrt{(x-9)^2+y^2} \\ \Rightarrow & 12 x-216 =-24 \sqrt{(x-9)^2+y^2} \\ \Rightarrow & x-18 =-2 \sqrt{(x-9)^2+y^2} \\ \Rightarrow & x^2-36 x+324 =4\left(x^2-18 x+81+y^2\right) \\ \Rightarrow & 3 x^2+4 y^2-36 x =0 \end{array} \end{aligned}$$

Let the point be $P(x, y)$.

$\therefore$ Distance from $(0,4)=\sqrt{x^2+(y-4)^2}$

So, the distance from the line $y=9$ is $\left|\frac{y-9}{\sqrt{1}}\right|$

$$\begin{array}{lr} \therefore & \sqrt{x^2+(y-4)^2}=\frac{2}{3}\left|\frac{y-9}{1}\right| \\ \Rightarrow & x^2+y^2-8 y+10=\frac{4}{9}\left(y^2-18 y+81\right) \\ \Rightarrow & 9 x^2+9 y^2-72 y+144=4 y^2-72 y+324 \\ \Rightarrow & 9 x^2+5 y^2=180 \end{array}$$

Let the points be $P(x, y)$.

$\therefore$ Distance of $P$ from $(4,0) \sqrt{(x-4)^2+y^2}\quad \text{... (i)}$

and the distance of $P$ from $(-4,0) \sqrt{(x+4)^2+y^2}\quad \text{... (ii)}$

Now,

$$\begin{aligned} & \sqrt{(x+4)^2+y^2}-\sqrt{(x-4)^2+y^2}=2 \\ & \sqrt{(x+4)^2+y^2}=2+\sqrt{(x-4)^2+y^2} \end{aligned}$$

On squaring both sides, we get

$$\begin{array}{rlrl} & x^2+8 x+16+y^2 =4+x^2-8 x+16+y^2+4 \sqrt{(x-4)^2+y^2} \\ \Rightarrow & 16 x-4 =4 \sqrt{(x-4)^2+y^2} \\ \Rightarrow & 4(4 x-1) =4 \sqrt{(x-4)^2+y^2} \\ \Rightarrow & 16 x^2-8 x+1=x^2+16-8 x+y^2 \\ \Rightarrow & 15 x^2-y^2 =15 \text { which is a parabola. } \end{array}$$

$$\begin{aligned} &\text { (i) Given that, vertices }=( \pm 5,0) \text {, foci }=( \pm 7,0) \text { and } a= \pm 5\\ &\begin{aligned} & \therefore \quad( \pm a e, 0)=( \pm 7,0) \\ & \text { Now } \quad a e=7 \Rightarrow 5 e=7 \\ & \Rightarrow \quad e=7 / 5 \\ & \because \quad b^2=a^2\left(e^2-1\right) \\ & \Rightarrow \quad b^2=25\left(\frac{49}{25}-1\right) \\ & \Rightarrow \quad b^2=25\left(\frac{49-25}{25}\right) \\ & \Rightarrow \quad b^2=24 \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { So,the equation of parabola is }\\ &\frac{x^2}{25}-\frac{y^2}{24}=1 \quad\left[\because a^2=25 \text { and } b^2=24\right] \end{aligned}$$

$$\begin{aligned} & \text { (ii) Vertices }=(0, \pm 7), e=4 / 3 \\ & \therefore \quad b=7, e=4 / 3 \\ & \because \quad e^2=1+\frac{a^2}{b^2} \\ & \Rightarrow \quad \frac{16}{9}-1=\frac{a^2}{49} \\ & \Rightarrow \quad \frac{7}{9}=\frac{a^2}{49} \Rightarrow a^2=\frac{343}{9} \end{aligned}$$

$$\begin{aligned} &\text { So, the equation of hyperbola is }\\ &\begin{aligned} & -\frac{x^2 \times 9}{343}+\frac{y^2}{49} =1 \\ \Rightarrow \quad & -\frac{9 x^2}{7}+y^2 =49 \\ \Rightarrow \quad & 9 x^2-7 y^2+343 =0 \end{aligned} \end{aligned}$$

(iii) Given that, foci $=(0, \pm \sqrt{10})$

$$\begin{array}{lr} \because & b e=\sqrt{10} \\ \Rightarrow & a^2+b^2=10 \\ \Rightarrow & a^2=10-b^2 \end{array}$$

$\therefore$ Equation of the hyperbola be

$$-\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\quad \text{.... (i)}$$

Since, this hyperbola passes through the point $(2,3)$.

$$\begin{array}{ll} \therefore & -\frac{4}{a^2}+\frac{9}{b^2}=1 \\ \Rightarrow & \frac{-4}{10-b^2}+\frac{9}{b^2}=1 \end{array}$$

$$\begin{array}{lr} \Rightarrow & \frac{-4 b^2+90-9 b^2}{b^2\left(10-b^2\right)}=1 \\ \Rightarrow & -13 b^2+90=10 b^2+b^4 \\ \Rightarrow & b^4-23 b^2+90=0 \\ \Rightarrow & b^4-18 b^2-5 b^2+90=0 \\ \Rightarrow & b^2\left(b^2-18\right)-5\left(b^2-18\right)=0 \\ \Rightarrow & \left(b^2-18\right)\left(b^2-5\right)=0 \\ \Rightarrow & b^2=18 \Rightarrow b= \pm 3 \sqrt{2} \\ \text { or } & b^2=5 \Rightarrow b=\sqrt{5} \\ \therefore & b^2=18 \text { then } a^2=-8 \quad \text{[not possible]}\\ \text { When } & a^2=5 \text {, then } b^2=5 \end{array}$$

$$\begin{aligned} &\text { So, the equation of hyperbola is }\\ &\begin{aligned} -\frac{x^2}{5}+\frac{y^2}{5} & =1 \\ y^2-x^2 & =5 \end{aligned} \end{aligned}$$