# Thread: Distance from point to line on hyperbola asymptotes problem

1. ## 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

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

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.

Stonehambey

2. Hi

Originally Posted by Stonehambey
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

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

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

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

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

Then you can calculate $\displaystyle \left(x^2-y^2\right)^2$ and $\displaystyle 4x^2(x^2-a^2)$ to show that they are equal

3. I also had trouble with messy algebra, but I got there in the end, using a more geometric/trigonometric approach.

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

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

If the point (x,y) also lies on the hyperbola then $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{m^2a^2} = 1$, from which $\displaystyle m^2(x^2-a^2) = y^2$. Substitute the formula for m from the previous paragraph into that, and you get the result.

4. 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

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

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

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

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

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

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