A fixed point $F(f,g)$ is at a perpendicular distance $a$ from a fixed straight line: $a_1x+b_1y+c=0$ and a point moves so that is distance from the fixed point is always equal to its distance from the fixed line. Find the equation to its locus,the axes of coordinates being drawn through the fixed point and being parallel and perpendicular to the given line.

We can write $$a= {a_1f+b_1g+c\over \sqrt{a_1^2+b_1^2}}$$

How to proceed further?


Hint: Suppose that the point is at origin and the line is $x=a$. Then each $(x,y)$ satisfying this condition should satisfy the following: $$ x^2+y^2=(x-a)^2 \implies y^2=a^2-2ax $$ You can find the similar equation for a general point and line by rotation and translation of the curve.

  • $\begingroup$ is it coincidence that the answer is x^2+2ay=a^2 $\endgroup$ Sep 20 '13 at 2:37
  • $\begingroup$ What do you mean by coincidence? The answer is the conclusion of the assumptions we had. $\endgroup$
    – Arash
    Sep 20 '13 at 14:10
  • $\begingroup$ there is no rotation constants or sumthng like that.. $\endgroup$ Sep 23 '13 at 19:25

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