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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.

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closed as off-topic by Bobson Dugnutt, Michael Hoppe, Harish Chandra Rajpoot, Shailesh, Stefan Mesken Mar 8 '16 at 0:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Bobson Dugnutt, Michael Hoppe, Harish Chandra Rajpoot, Shailesh, Stefan Mesken
If this question can be reworded to fit the rules in the help center, please edit the question.

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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:

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

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

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

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  • $\begingroup$ For searching purposes: the resulting curve is called an astroid. $\endgroup$ – J. M. is a poor mathematician Sep 4 '17 at 17:57

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