# What is the equation of the parabola that has a vertex at  (14, -9)  and passes through point  (0, -5) ?

Feb 1, 2017

See explanation, for the existence of a family of parabolas
Upon imposing one more condition that that the axis is x-axis, we get a member $7 {y}^{2} - 8 x + 70 y + 175 = 0$.

#### Explanation:

From definition of the parabola, the general equation to a parabola

having focus at $S \left(\alpha , \beta\right)$ and directrix DR as y=mx+c is

$\sqrt{{\left(x - \alpha\right)}^{2} + {\left(y - \beta\right)}^{2}} = | y - m x - c \frac{|}{\sqrt{1 + {m}^{2}}}$,

using 'distance from S = distance from DR'.

This equation has $4$ parameters $\left\{m , c , \alpha , \beta\right\}$.

As it passes through two points, we get two equations that relate

the $4$ parameters.

Of the two points, one is the vertex that bisects the perpendicular

from S to to DR, $y - \beta = - \frac{1}{m} \left(x - \alpha\right)$. This gives

one more relation. The bisection is implicit in the already obtained

equation. Thus, one parameter remains arbitrary. There is no unique

solution.

Assuming that the axis is x-axis, the equation has the form

${\left(y + 5\right)}^{2} = 4 a x$. This passes through $\left(14 , - 9\right)$.

So, $a = \frac{2}{7}$ and equation becomes

$7 {y}^{2} - 8 x + 70 y + 175 = 0.$

Perhaps, a particular solution like this is required.