# What is the equation for a parabola with vertex:(8,6) and focus:(3,6)?

Jul 19, 2016

For the parabola it is given

$V \to \text{Vertex} = \left(8 , 6\right)$

$F \to \text{Focus} = \left(3 , 6\right)$

We are to find out the equation of the parabola

The ordinates of V(8,6) and F(3,6) being 6 the axis of parabola will be parallel to x-axis and its equation is $y = 6$

Now let the coordinate of the point (M) of intersection of directrix and axis of parabola be $\left({x}_{1} , 6\right)$.Then V will be mid point of MF by the property of parabola. So

$\frac{{x}_{1} + 3}{2} = 8 \implies {x}_{1} = 13$
$\text{Hence} M \to \left(13 , 6\right)$

The directrix which is perpendicular to the axis ($y = 6$) will have equation $x = 13 \mathmr{and} x - 13 = 0$

Now if$P \left(h , k\right)$ be any point on the parabola and N is the foot of the perpendicular drawn from P to the directrix,then by the property of parabola

$F P = P N$

$\implies \sqrt{{\left(h - 3\right)}^{2} + {\left(k - 6\right)}^{2}} = h - 13$

$\implies {\left(h - 3\right)}^{2} + {\left(k - 6\right)}^{2} = {\left(h - 13\right)}^{2}$

$\implies {\left(k - 6\right)}^{2} = {\left(h - 13\right)}^{2} - {\left(h - 3\right)}^{2}$

=>(k^2-12k+36=(h-13+h-3)(h-13-h+3)

$\implies {k}^{2} - 12 k + 36 = \left(2 h - 16\right) \left(- 10\right)$

$\implies {k}^{2} - 12 k + 36 + 20 h - 160 = 0$

$\implies {k}^{2} - 12 k + 20 h - 124 = 0$

Replacing h by x and k by y we get the required equation of the parabola as

$\textcolor{red}{{y}^{2} - 12 y + 20 x - 124 = 0}$