Hi,

I've been struggling with this for a while now and to no avail. Here's the question.

A hyperbola of the form

$\displaystyle \frac{x^2}{\alpha^2}-\frac{y^2}{\beta^2}=1$

has asymptotes with equations $\displaystyle y^2 = m^2 x^2$ and passes through the point (a,0). Find the equation of the hyperbola in terms of x, y, a and m.

A point P on this hyperbola is equidistant from one of its asymptotes and the x-axis. Prove that, for all values of m, P lies on the curve with equation.

$\displaystyle (x^2-y^2)^2 = 4x^2(x^2-a^2)$

I hate these type of questions as I always end up with a pile of messy algebra and ton of screwed up paper. Here's my attempt this time.

I got the first part ok, the equation of the hyperbola is

$\displaystyle \frac{x^2}{a^2}-\frac{y^2}{m^2 a^2}=1$

For the second part I decided to pick an arbitrary point on the hyperbola and find its distance from the asymptote and the x-axis and set them equal, then eliminate m. So my general point is $\displaystyle P = (a\sec\theta, am\tan\theta)$. Now the distance of this point to the x-axis is simply $\displaystyle am\tan\theta$. For the moment I chose the point to be in the first quadrant so the equation of the corresponding asymptote is $\displaystyle y=mx$. Wolfram says that the distance between the point and the asymptote is

$\displaystyle \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$ where the line is $\displaystyle Ax + By + C = 0$

So for my point and line this becomes

$\displaystyle \frac{ma\sec\theta + ma\tan\theta}{\sqrt{m^2+1}}$ which I set equal to the distance from the y-axis to obtain

$\displaystyle \frac{ma \sec \theta + ma\tan\theta}{\sqrt{m^2+1}} = ma\sec\theta$

It is here where I have no idea how to eliminate m to get the required equation or even if I've been doing it wrong the whole time.

Please can someone help, thanks!!

Stonehambey