Parabola general equation

Question

how to get the general equation of the following parabola

(a) one whose vertex is on y = 2, with latus rectum of length 6, and contains (2, 8). (b) one whose vertex is on the y-axis, symmetry axis parallel to x, and contains (8, −1).

Answer 1

The general equation of any conic section is

$$ax^2+by^2+cxy+dx+ey+f=0$$

For a parabola you have $ab-c^2=0$. This is also called the discriminant of the conic. Multiples of this equation describe the same conic, as the equation is homogenic. So this leaves you with 4 real degrees of freedom which have actual geometric meaning. 6 degrees for the 6 variables minus 1 degree for homogenity minus 1 degree for parabola condition.

Both your setups constrain 3 degrees of freedom. Knowing a coordinate of the vertex, or knowing the direction of the symmetry axis, or knowing the coordinates of one point on the conic section, or knowing the latus rectum, each costs you one degree of freedom. Which means that for both your scenarios you'd expect a one-parameter family of parabolas. Thus your general equation would likely be in terms of the coordinates $x$, $y$ and some parameter $t$.

Of the two problems, (b) is the easier one. Simply pick the $y$ coordinate of the vertex as your parameter $t$. So you have a parabola with vertex at $(0,t)$ and symmetry axis in the $x$ direction. So its equation has to be something like $x=k(y-t)^2$. From the given point $(8,-1)$ you know $8=k(1+t)^2$ so $k=\frac8{(1+t)^2}$. This lets you rewrite $x=k(y-t)^2$ as $(1+t)^2x=8(y-t)^2$ or eventually

$$0x^2-8y^2+0xy+(1+t)^2x+16ty-8t^2=0$$

from which you can read the coefficients $a$ through $f$ if you want to.

Finding a “nice” parametrization for (a) is tricky. For example, if you pick the $x$ coordinate of the vertex as your parameter, this still leaves you with two choices for each vertex position, depending on which branch the given point is on. A parametrization which avoids this issue might be preferable.

Barring better ideas, you can try to find the equation of any parabola with the given latus rectum, and choose a generic point on that line (i.e. parametrized by $t$). Then find an isometric transformation which maps the generic point to the given point, and the vertex of the chosen parabola to a point on the given line where the final vertex should be. Picking this isometry involves finding a point on the given line, a given distance away from the given point. So it's a circle-line intersection, which is where the two choices I described come into play.

I'm still experimenting in the hope of finding a nicer parametrization.