Question

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

Answer 1

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 `7y^2-8x+70y+175=0`.

Explanation:

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

having focus at `S(alpha, beta)` and directrix DR as y=mx+c is

`sqrt((x-alpha)^2+(y-beta)^2)=|y-mx-c|/sqrt(1+m^2)`,

using 'distance from S = distance from DR'.

This equation has `4` parameters `{m, c, alpha, beta}`.

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=-1/m(x-alpha)`. 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

`(y+5)^2=4ax`. This passes through `(14, -9)`.

So, `a = 2/7` and equation becomes

`7y^2-8x+70y+175=0.`

Perhaps, a particular solution like this is required.