 # If the points (0, 4) and (0, 2) are respectively `
Question:

If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then find the equation of the parabola.

Solution:

Given

Vertex $=(0,4)$ Focus $=(0,2)$

So, the directrix of the parabola is $y=6$,

Since, distance of $(x, y)$ from $(0,2)$ and perpendicular distance from $(x, y)$ to directrix are always equal.

Using Distance Formula \& Perpendicular Distance Formula, Perpendicular Distance (Between a point and line) is

$=\frac{\left|a x_{1}+b y_{1}+c\right|}{\sqrt{a^{2}+b^{2}}}$

Where the point is $\left(x_{1}, y_{1}\right)$ and the line is expressed as ax $+$ by $+c=0$ i.e.., $x(0)+y-6=0 \&(x, y)$

Distance between the point of intersection \& centre

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

Now by substituting the values we get

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

Squaring both the sides,

$\left[\sqrt{\left((x-0)^{2}+(y-2)^{2}\right)}\right]^{2}=(y-6)^{2}$

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

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

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