Find the focus of a parabola that touches $x=0$, $y=0$, $x-y+1=0$, and $-2x-y-8=0$.

Question

Find the focus of a parabola that touches $x=0$, $y=0$, $x-y+1=0$, and $-2x-y-8=0$.

I am not getting how to find the equation of parabola from these four tangents.

Can anyone help in this?

Answer 1

My first inclination was to compute an equation of the parabola using essentially Jan-Magnus Økland’s method, but the problem posed in the quote block is to find its focus, and this can be done directly, without finding the parabola itself.

The foot of the perpendicular from a parabola’s focus to any tangent line lies on the tangent to its vertex. So, if the coordinates of the focus are $(h,k)$, we know from the fact that the parabola is tangent to the coordinate axes that an equation of this line is $x/h+y/k=1$. Computing the feet of the perpendiculars to the other two tangents and requiring that they lie on this line produces the following system of equations: $${h^2+2hk+k^2+h-k\over2hk}=1 \\ {2h^2-5hk+2k^2+8h+16k\over5hk}=-1$$ with solution $h=-6/5$, $k=2/5$.

If you do need an equation for this parabola, one is easily developed using this focus. We have a pair of perpendicular tangents that intersect at the origin, so we know the directrix passes through the origin. It is also parallel to the vertex tangent found above, so with a little rearrangement we can find the equation $x-3y=0$ for it. Using the formulas for distance to a point and a line, an equation of the parabola is therefore $${(x-3y)^2\over10}=\left(x+\frac65\right)^2+\left(y-\frac25\right)^2.$$ Rearrange as desired.

Answer 2

All conics have the form $ax^2+2bxy+cy^2+dx+ey+f=0$.
A parabola has $ac=b^2$.
If it touches $x=0$ then, when you put $x=0$ in, the result is a quadratic in $y$ that is a perfect square. And so with the other lines.

Answer 3

The unique conic tangent to five lines can be found by dualizing, taking the usual determinant to find the conic through five points and dualizing back.

Finding the conic through $(1:-1:1),(1/4:1/8:1),(1:0:0),(0:1:0),(0:0:1)$ is to take the determinant

$\begin{pmatrix} x^{2}&x\,y&y^{2}&x\,z&y\,z&z^{2}\\ 1&0&0&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&0&0&1\\ 1&{-1}&1&1&{-1}&1\\ \frac{1}{16}&\frac{1}{32}&\frac{1}{64}&\frac{1}{4}&\frac{1}{8}&1\end{pmatrix}=-\frac3{32}(4xy + xz - 3yz)$

These points correspond to the five lines your conic is tangent to: $$\begin{align}(1:-1:1)&\mapsto x-y+1=0\\ (\frac14:\frac18:1)&\mapsto \frac{x}{4}+\frac{y}{8}+1=0\\ (1:0:0)&\mapsto x=0\\ (0:1:0)&\mapsto y=0\\ (0:0:1)&\mapsto z=0\end{align}$$ The fifth being the line at infinity since it's a parabola.

Now the last part is finding the dual conic. The dual of a conic is found by taking the adjugate (or inverse if possible) of the corresponding symmetric matrix. $$N=\begin{pmatrix} 0&2&\frac{1}{2}\\ 2&0&{-\frac{3}{2}}\\ \frac{1}{2}&{-\frac{3}{2}}&0\end{pmatrix}$$ and taking the inverse of this.

$$N^{-1}=\begin{pmatrix} \frac{3}{4}&\frac{1}{4}&1\\ \frac{1}{4}&\frac{1}{12}&{-\frac{1}{3}}\\ 1&{-\frac{1}{3}}&\frac{4}{3}\end{pmatrix}$$

This is the conic in the projective plane (well, I multiplied the equation by $12$ to clear denominators): $9\,x^{2}+6\,x\,y+y^{2}+24\,x\,z-8\,y\,z+16\,z^{2}=0$ . Setting $z=1$ gives the parabola in the plane: $$9x^2+6x y+y^2+24x-8y+16=0.$$

The focus can be read off of the following form of the same equation $$10((x+6/5)^2+(y-2/5)^2-(3y-x)^2/10)=0.$$

Answer 4

Here's one more approach which I believe is very simple:

As you already know that we have two pairs of perpendicular tangents which intersect on the directrix, The equation of directrix is readily known.

Now assume focus to be $(h,k)$. The reflection of the focus over any tangent lies on the directrix which results in that all you have to solve is a system of linear equations to find the focus.