To find the locus of vertices of shifted parabola

Question

How to deal with this question. Please help.

"Parabolas are drawn to touch two given rectangular axes and their foci are all at a constant distance $c$ from the origin. Find the locus of the vertices of the parabola."

This question seems like the shifting of parabola to different axes. The general equation of parabola is $(ax+by)^2+2gx+2fy+c=0$. The $x-axis$ and the $y-axis$ will be the tangents of the parabola.

Answer 1

Useful facts

1. Perpendicular tangents meet at the directrix.
2. As in figure 1 $F$ and $C$ is symmetrical about the tangent line $BE$. $C$ is the perpendicular foot on the directrix.

Let $F=(c\cos \theta, c\sin \theta)$, then the directrix is

$$\frac{y}{x}=-\tan \theta$$

$$x\sin \theta+y\cos \theta=0 \quad \cdots \cdots (1)$$

By $EF=EC$, the equation of parabola is

$$(x-c\cos \theta)^{2}+(y-c\sin \theta)^{2}=(x\sin \theta+y\cos \theta)^{2}$$

$$(x\cos \theta-y\sin \theta)^{2}-2c(x\cos \theta+y\sin \theta)+c^{2}=0$$

which touches the axes at $(c\sec \theta,0)$ and $(0,c\csc \theta)$.

Equation of the principal axis:

$$\frac{y-c\sin \theta}{x-c\cos \theta}=\cot \theta $$

$$x\cos \theta-y\sin \theta = c\cos 2\theta \quad \cdots \cdots (2)$$

$(1) \cap (2) \implies \left ( \begin{array}{c} x \\ y \end{array} \right)= \left ( \begin{array}{c} c\cos \theta \cos 2\theta \\ -c\sin \theta \cos 2 \theta \end{array} \right)$

The locus of the vertex is:

$$2\left ( \begin{array}{c} x \\ y \end{array} \right)= \left ( \begin{array}{c} c\cos \theta \\ c\sin \theta \end{array} \right)+ \left ( \begin{array}{c} c\cos \theta \cos 2\theta \\ -c\sin \theta \cos 2 \theta \end{array} \right)$$

$$\left ( \begin{array}{c} x \\ y \end{array} \right)= \left ( \begin{array}{c} c\cos^{3} \theta \\ c\sin^{3} \theta \end{array} \right)$$

$$x^{2/3}+y^{2/3}=c^{2/3}$$