# Hyperbola-locus of P equidist from asymptotes and x-axis?

• February 27th 2009, 05:28 AM
Hyperbola-locus of P equidist from asymptotes and x-axis?
A hyperbola of the form
$\frac{x^2}{\alpha}-\frac{y^2}{\beta}=1$
has asymptotes with equation $y^2=m^2 x^2$ and passes through the point (a,0). Find an equation of the hyperbola in terms of x, y, a and m.
Ans:
$\frac{x^2}{a^2}-\frac{y^2}{a^2 m^2}=1$
A point P on this hyperbola is equidistant from one of its asymptoes and the x-axis. Prove that, for all values of m, P lies on the curve with equation
$(x^2-y^2)^2=4x^2 (x^2-a^2)$
• February 27th 2009, 04:33 PM
ThePerfectHacker
Quote:

A hyperbola of the form
$\frac{x^2}{\alpha}-\frac{y^2}{\beta}=1$
has asymptotes with equation $y^2=m^2 x^2$ and passes through the point (a,0). Find an equation of the hyperbola in terms of x, y, a and m.
Ans:
$\frac{x^2}{a^2}-\frac{y^2}{a^2 m^2}=1$
A point P on this hyperbola is equidistant from one of its asymptoes and the x-axis. Prove that, for all values of m, P lies on the curve with equation
$(x^2-y^2)^2=4x^2 (x^2-a^2)$
The asymptotes of $\frac{x^2}{\alpha} - \frac{y^2}{\beta} = 1$ can be obtained by solving $\frac{x^2}{\alpha} - \frac{y^2}{\beta} = 0 \implies y^2 = \frac{\beta}{\alpha}x^2$.
Therefore, $m^2 = \frac{\beta}{\alpha} \implies \beta = \alpha m^2$.
Now the hyperbola passes through $(a,0)$, therefore $\frac{a^2}{\alpha} = 1 \implies \alpha = a^2 \implies \beta = a^2 m^2$.
The $P = (x_0,y_0)$ lie in the first quadrant, and the asympote be $y=mx$.
The distance between $P$ and the asymptote is $\frac{|y_0 - mx_0|}{\sqrt{1+m^2}}$.
The distance between $P$ and the x-axis is $x_0$.
Thus, we want, $\frac{|y_0 - mx_0|}{\sqrt{1+m^2}} = x_0 \implies (y_0 - mx_0)^2 = x_0^2 (1+m^2)$.