# edexcel p5 parabola question

Watch
Announcements
This discussion is closed.
#1
A hyperbola of the form
x^2/(alpha)^2 - y^2/β^2 =1
has asymptotes with equation y^2=m^2x^2 and passes through the point (a,0). Find an euqation 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)
0
15 years ago
#2
(Original post by totaljj)
A hyperbola of the form
x^2/(alpha)^2 - y^2/β^2 =1
has asymptotes with equation y^2=m^2x^2 and passes through the point (a,0). Find an euqation 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)
For the first bit I get x²/a² - y²/a²m² = 1

Not too sure about the second bit... thought about doing it in parameters, so considering a point (a sect, m/a tant) then using the formula for the closest distance of a point to a line (which may be in the formula book) - the line being y = mx ( or maybe y = -mx it shouldn't matter) and equating it to m/a tant.
0
15 years ago
#3
Okay for the second bit first write the equation in terms of parameters, that is x = a sect and y = ma tant

Now the distance to the x-axis is just the y coordinate ie ma tant. The perpendicular distance of a point to a line is in the formula book so the distance of (a sect, ma tant) to mx - y = 0 is...

(ma sect - matant)/sqrt(1 + m²) and from the question this is equal to ma tant. Squaring gives...

(sect - tant)² = (1 + m²)tan²t

ie sec²t - 2sect.tant = m²tan²t

putting back in (x,y) gives x²/a² - 2xy/ma² = y²/a²

Now the question asks for the equation for all m so we must find an expression independant of m. To do this rearrange the above to get m in terms of x, y and a...

2xy/m = x² - y² so 1/m = (x² - y²)/2xy

But the point lies on the hyperbola, so x² - y²/m² = a²

sub in 1/m² gives x² - y²((x² - y²)/2xy)² = a²

so 4x²y²x² - y²(x² - y²)² = 4a²x²y² cancelling y² and factorising gives...

4x²(x² - a²) = (x² - y²)²

Hope that helps (even if it is a bit late!)
0
#4
thank you phill
0
X
new posts Back
to top

view all
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

Yes, my partner and I are struggling (28)
7.87%
Yes, my partner and I broke up (32)
8.99%
Yes, it's hard being around my family so much (78)
21.91%
Yes, I'm feeling lonely isolating alone (44)
12.36%
No, nothing has changed (109)
30.62%
No, it's helped improve my relationships (65)
18.26%