Distance from point to line on hyperbola asymptotes problem

**Stonehambey** · August 9th 2009, 08:25 AM

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.

Please can someone help, thanks!!

Stonehambey

**running-gag** · August 9th 2009, 10:43 AM

Hi

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

**Opalg** · August 9th 2009, 11:33 AM

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.

**Stonehambey** · August 9th 2009, 11:53 AM

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

Thread: Distance from point to line on hyperbola asymptotes problem

LinkBack

Thread Tools

Display

Distance from point to line on hyperbola asymptotes problem

Similar Math Help Forum Discussions

Curve's point A tangent line passes through a point P. What is the distance of AP

distance of a point from a line

distance of line and a point

Calculating coords of a point on a straight line at a distance from start point...

distance from point to line

Search Tags