Question

Find the equation of the following parabola

A) $\text{Directrix},x=0,\text{focus at}(6,0)$

B) $\text{Vertex at}(0,4),\text{focus at}(0,2)$

C) $\text{Focus at}(-1,-2),\text{directrix}x-2y+3=0$

Answer 1

A)

Given: Focus $S(6,0)$, Directrix: $x=0$
From diagram, it is clear that directrix passes through $O(0,0)$.
Vertex $(3,0)$
$(\because$ Vertex is mid-point of Focus and $O(0,0))$
Parabola is locus of points whose distance from a fixed point (Focus) and from a fixed line (Directrix) are equal.
$\Rightarrow PS=PM$

Perpendicular distance of point
$(x_1,y_1)$ from line $ax+by+c=0$
is $\frac{|a{x}_1+b{y}_1+c|}{\sqrt{\surd a^2+b^2}}$

$\Rightarrow \sqrt{(x-6{)}^2+(y-0{)}^2}=|x|$

Squaring on both sides
$\Rightarrow x^2+36-12x+y^2=x^2$
$\Rightarrow y^2-12x+36=0$

Hence, equation of required parabola is $y^2-12x+36=0$

B)


From the diagram, it’s clear that, directrix is $y=6$
Or $y-6=0$
Parabola is locus of points whose distance from a fixed point (Focus) and from a fixed line (Directrix) are equal.
$\Rightarrow PS=PM$
$\left[\text{Given, Focus}S\right(0,2\left)\right]$

Perpendicular distance of point
$(x_1,y_1)$ from line $ax+by+c=0$
is $\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$

$\Rightarrow \sqrt{(x-0{)}^2+(y-2{)}^2=|y-6|}$

Squaring on both sides, we have
$x^2+y^2+4-4y=y^2+36-12y$
$\Rightarrow x^2+8y=32$

Hence, equation of required parabola is $x^2+8y=32$

C)


Given, Focus at $(-1,-2)$
And directrix is $x-2y+3=0$
Parabola is locus of points whose distance from a fixed point (Focus) and from a fixed line (Directrix) are equal.
$\Rightarrow PS=PM$
$\left[\text{Given, Focus}S\right(-1,-2\left)\right]$

Perpendicular distance of point
$(x_1,y_1)$ from line $ax+by+c=0$is
$\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$

$\Rightarrow \sqrt{(x+1{)}^2+(y+2{)}^2}=\frac{|x-2y+3|}{\sqrt{1+4}}$
Squaring on both sides, we get
$x^2+1+2x+y^2+4+4y$

$=\frac{x^2+4y^2+9-4xy-12y+6x}{5}$

$\Rightarrow 4x^2+4xy+y^2+4x+32y+16=0$

Hence, equation of required parabola is $4x^2+4xy+y^2+4x+32y+16=0$.