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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.

Pratyush
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1 Answers1

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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:

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

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

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

Ng Chung Tak
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