Find equation of hyperbola, symmetric about the origin, traced by a point whose distance to point $(4,0)$ is twice its distance to line $x=1$.

Question

The question is as follows:

Find the equation of a hyperbola which is symmetrical about the origin traced by a point that moves so its distance from the point $(4,0)$ is twice as far as the point is from the line $x=1$. Find the equation of the asymptotes of the curve.

I understand that the hyperbola is centered at the origin, but I need to find $b$ and $a$, so that I have the gradient for the equation of the asymptotes, but I'm unsure how to go about finding this. The wording really confuses me.

Any help or possible explanation concerning this would be of great help. Thanks in advance. :)

Answer 1

The point that the question refers to is not fixed; it moves. Every point $(x,y)$ on the hyperbola must satisfy the condition that $(x,y)$ is twice as far from $(4,0)$ as it is from $(1,y)$. This translates to the condition $$ \sqrt{(x-4)^2+y^2} = 2 \sqrt{(x-1)^2}. $$ If you square both sides and rearrange, you should get the equation of a hyperbola in the form $$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1. $$ Then the asymptotes will be the lines $y = \pm \frac{b}{a}x$.