Hi,
I've been struggling with this for a while now and to no avail. Here's the question.
A hyperbola of the form
has asymptotes with equations 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.
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
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 . Now the distance of this point to the x-axis is simply . For the moment I chose the point to be in the first quadrant so the equation of the corresponding asymptote is . Wolfram says that the distance between the point and the asymptote is
where the line is
So for my point and line this becomes
which I set equal to the distance from the y-axis to obtain
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
I also had trouble with messy algebra, but I got there in the end, using a more geometric/trigonometric approach.
The asymptote makes an angle with the x-axis, where . If a point is equidistant from these two lines, then it must lie on their bisector, namely the line , where . The formula for tan of twice an angle then tells you that .
Therefore . If the point (x,y) lies on the line then , and so , from which .
If the point (x,y) also lies on the hyperbola then , from which . Substitute the formula for m from the previous paragraph into that, and you get the result.
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
Now I square again because this remove any ambiguity as to which asymptote we are referring to, so we now have
Now, I used the fact that x and y lie on a hyperbola to get
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