**Question:**

**Find the equation of each of the following parabolas**

**(a) Directrix x = 0, focus at (6, 0)**

**(b) Vertex at (0, 4), focus at (0, 2)**

**(c) Focus at (–1, –2), directrix x – 2y + 3 = 0**

**Solution:**

(a) The distance of any point on the parabola from its focus and its directrix is same.

Given that, directrix, x = 0 and focus = (6, 0)

If a parabola has a vertical axis, the standard form of the equation of the parabola is (x – h)2 = 4p (y – k), where p≠ 0.

The vertex of this parabola is at (h, k).

The focus is at (h, k + p) & the directrix is the line y = k – p.

As the focus lies on x – axis,

Therefore the equation is y2 = 4ax or y2 = -4ax

So, for any point P(x, y) on the parabola

Distance of point from directrix = Distance of point from focus

x2 = (x – 6)2 + y2

x2 = x2 – 12x + 36 + y2

y2 – 12x + 36 = 0

Hence the required equation is y2 – 12x + 36 = 0.

(b) Given Vertex $=(0,4) \&$ Focus $=(0,2)$

We know the distance between the vertex and directrix is same as the distance

between the vertex and focus.

Directrix is $y-6=0$

For any point of $P(x, y)$ on the parabola

Distance of $P$ from directrix $=$ Distance of $P$ from focus

Perpendicular Distance (Between a point and line) $=\frac{\left|a x_{1}+b y_{1}+c\right|}{\sqrt{a^{2}+b^{2}}}$

Where the point is $\left(x_{1}, y_{1}\right)$ and the line is expressed as $a x+b y+c=0$ i.e..,

$x(0)+y-6=0 \&$ point $=(x, y)$

Distance between the point of intersection \& centre

Distance formula $=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}}$ (between $(x, y) \&(0,2))$

$\frac{|0(x)+(1) y-6|}{\sqrt{0^{2}+1^{2}}}=\sqrt{\left((x-0)^{2}+(y-2)^{2}\right)}$

$\frac{y-6}{1}=\sqrt{x^{2}+y^{2}-4 y+4}$

Squaring both the sides,

$x^{2}+y^{2}-4 y+4=(y-6)^{2}$

$x^{2}+y^{2}-4 y+4=y^{2}-12 y+36$

$x^{2}+8 y-32=0$

Hence, the required equation is $x^{2}+8 y-32=0$.

(c) Focus $=(-1,-2)$, directrix is $x-2 y+3=0$

For any point $(x, y)$ on parabola, the distance from focus to that point is always equal to the perpendicular distance from that point to the directrix,

Perpendicular Distance (Between a point and line) $=\frac{\left|a x_{1}+b y_{1}+c\right|}{\sqrt{a^{2}+b^{2}}}$

Where the point is $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ and the line is expressed as $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$ i.e..,

$x-2 y+3=0$ and point $=(x, y)$

Distance between the point of intersection \& centre

Distance formula $=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$\frac{|1(x)+(-2) y+3|}{\sqrt{(-2)^{2}+1^{2}}}=\sqrt{\left((x-(-1))^{2}+\left(y-(-2)^{2}\right)\right.}$

$\frac{|x-2 y+3|}{\sqrt{4+1}}=\sqrt{\left((x+1)^{2}+(y+2)^{2}\right)}$

$\frac{x-2 y+3}{\sqrt{5}}=\sqrt{\left((x+1)^{2}+(y+2)^{2}\right)}$

Squaring both the sides,

$\left[\frac{x-2 y+3}{\sqrt{5}}\right]^{2}=\left[\sqrt{\left((x+1)^{2}+(y+2)^{2}\right)}\right]^{2}$

$\frac{(x-2 y+3)^{2}}{5}=(x+1)^{2}+(y+2)^{2}$

$x^{2}+4 y^{2}+9-4 x y+6 x-12 y=5\left[x^{2}+2 x+1+y^{2}+4 y+4\right]$

$x^{2}+4 y^{2}+9-4 x y+6 x-12 y=5 x^{2}+10 x+5 y^{2}+20 y+25$

$4 x^{2}+y^{2}+4 x y+4 x+32 y+16=0$

Hence the required equation is $4 x^{2}+y^{2}+4 x y+4 x+32 y+16=0$