I have problem in dealing the question related to the equation of shifted parabola. I have a question as
"A parabola whose latus rectum is $4c$, slide between two rectangular axes. Find the locus of focus and the equation of curve traced out by the parabola."
Using standard form of the parabola: $(x,y)=(ct^{2},2ct)$, then the equation of the tangent at $[t]$ is
$$x-ty+ct^{2}=0 \quad \ldots \ldots (1) $$
If tangents at $[t]$ and $[t']$ are perpendicular, $$tt'=-1$$
Then the equation of another tangent at $[t']$ is $$x+\frac{y}{t}+\frac{c}{t^{2}}=0$$
That is $$t^{2}x+ty+c=0 \quad \ldots \ldots (2) $$
The required locus is given by the perpendicular distances of the focus $(c,0)$ from the tangents:
\begin{align*} (X,Y) &= \left( \frac{c+0+ct^{2}}{\sqrt{1+t^{2}}}, \frac{t^{2}c+0+c}{\sqrt{t^{4}+t^{2}}} \right) \\ &= \left( c\sqrt{1+t^{2}}, \frac{c\sqrt{1+t^{2}}}{t} \right) \\ \end{align*}
Confine the sliding parabola in the first quadrant by taking $t>0$.
Considering $\dfrac{X}{Y}=t$ and eliminating $t$, we have
$$\frac{1}{X^{2}}+\frac{1}{Y^{2}}=c^{2}$$
Alternatively, let $t=\tan \theta$,
$$(X,Y)=(c\sec \theta, c\csc \theta) $$
