# Distance from point to line on hyperbola asymptotes problem

• Aug 9th 2009, 09:25 AM
Stonehambey
Distance from point to line on hyperbola asymptotes problem
Hi,

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

A hyperbola of the form

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

has asymptotes with equations $y^2 = m^2 x^2$ 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.

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

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

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

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 $P = (a\sec\theta, am\tan\theta)$. Now the distance of this point to the x-axis is simply $am\tan\theta$. For the moment I chose the point to be in the first quadrant so the equation of the corresponding asymptote is $y=mx$. Wolfram says that the distance between the point and the asymptote is

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

So for my point and line this becomes

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

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

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.

Stonehambey
• Aug 9th 2009, 11:43 AM
running-gag
Hi

Quote:

Originally Posted by Stonehambey
Wolfram says that the distance between the point and the asymptote is

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

So for my point and line this becomes

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

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

Yes but your equation is $y = mx$ or $mx - y = 0$ therefore $A = m$ and $B = -1$ which gives $\frac{ma\sec\theta - ma\tan\theta}{\sqrt{m^2+1}}$

Then I do not understand why you set it equal to the distance from the y-axis ? It should be the x-axis right ?

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

You can then eliminate $ma$ and simplify to get $m^2 = \frac{1-2\sin \theta}{\sin^2 \theta}$

Then you can calculate $\left(x^2-y^2\right)^2$ and $4x^2(x^2-a^2)$ to show that they are equal
• Aug 9th 2009, 12:33 PM
Opalg
I also had trouble with messy algebra, but I got there in the end, using a more geometric/trigonometric approach.

The asymptote $y=mx$ makes an angle $\theta$ with the x-axis, where $m=\tan\theta$. If a point is equidistant from these two lines, then it must lie on their bisector, namely the line $y=tx$, where $t = \tan(\theta/2)$. The formula for tan of twice an angle then tells you that $m = \frac{2t}{1-t^2}$.

Therefore $m(1-t^2) = 2t$. If the point (x,y) lies on the line $y=tx$ then $t = \tfrac yx$, and so $m\bigl(1-\bigl(\tfrac yx\bigr)^2\bigr) = 2\tfrac yx$, from which $m = \frac{2xy}{x^2-y^2}$.

If the point (x,y) also lies on the hyperbola then $\frac{x^2}{a^2} - \frac{y^2}{m^2a^2} = 1$, from which $m^2(x^2-a^2) = y^2$. Substitute the formula for m from the previous paragraph into that, and you get the result.
• Aug 9th 2009, 12:53 PM
Stonehambey
That's a pretty smart way! :)

I got there in the end by saying that the point be (x,y) and using their relationship only at the end, it went something like this

$y = \frac{mx+y}{\sqrt{m^2+1}}$
$y^2 m^2 + y^2 = m^2 x^2 + 2mxy + y^2$
$m^2(y^2-x^2) = 2mxy$

Now I square again because this remove any ambiguity as to which asymptote we are referring to, so we now have

$m^4 (y^2 - x^2)^2 = 4m^2 x^2 y^2$
$m^2 (y^2 - x^2)^2 = 4 x^2 y^2$

Now, I used the fact that x and y lie on a hyperbola to get

$m^2(x^2 - y^2)^2 = 4x^2 m^2 (x^2-a^2)$
$(x^2 - y^2)^2 = 4x^2(x^2 - a^2)$

So I got there in the end, but thank you for all your help :) I don't know what it is about these type of questions (conic section/locus ones) but they seem to knock me dead with insane algebra every single time. Are they generally quite tricky or am I being really dumb! :P

Stonehambey