The Student Room Group

Two Hyperbolae Questions

P5/FP2 - Ex4C

8) Prove that the triangle formed by the assymptotes to the curve with equation x2 - 2y2 = 4 and any tangent to the curve is of constant area.

Let P(c,d) be the point on the parabola where it meets the tangent.

I get the vertices of the triangle to be O(0,0) Q(√2c +d, c+ d/√2) and R(c- √2d, c- d/√2), but don't know where to go from here.


16) The variable point P(asect, btant) is on the hyperbola with equation x2/a2 - y2/b2 = 1 and N is the point (3a, 3b). The point Q lies on PN so that PQ = 2QN (PQ & QN are vectors). As t varies, find the locus of Q in cartesian form.

Any help is much appreciated.
Reply 1
8.
The equation of the tangent at P is:
xc - 2yd = 4

The asymptotes have equation y = ± (1/2) x, so that 2y^2 = x. Now suppose the asymptotes intersect the tangent:
(xc - 4)^2 = (2sqrt(x/2)d)^2
x^2 c^2 - 8xc + 16 = 2 x^2 d^2
x^2 (2d^2 - c^2) + 8cx - 16 = 0

Note that, from the equation of the curve, 2d^2 - c^2 = -(c^2 - 2d^2) = -4. Hence:
x^2 - 2cx + 4 = 0

Suppose the roots of this equation are r and s (they're the x-coords of the point of intersection of the tangent and the asymptotes), then rs = -4, which is constant. Since you showed that (0, 0) is one of the triangles vertices, this proves that the length of the other 2 sides of the triangle emerging from the origin is constant. So the triangle has constant area.

16.
Suppose Q has coords (x, y). Then:
PQ = (x - asec(t), y - btan(t))
QN = (3a - x, 3b - y)

So:
PQ = 2QN
(x - asec(t), y - btan(t)) = (6a - 2x, 6b - 2y)

x - asec(t) = 6a - 2x
=> sec(t) = (6a - 3x)/a

y - btan(t) = 6b - 2y
=> tan(t) = (6b - 3y)/b

Hence:
sec^2(t) - tan^2(t) = 1
(6a - 3x)^2/a^2 - (6b - 3y)^2/b^2 = 1
and simplify.
Reply 2
dvs
8.
The equation of the tangent at P is:
xc - 2yd = 4

The asymptotes have equation y = ± (1/2) x, so that 2y^2 = x.

Is it not y = ± (1/√2)x ?
Reply 3
yes.
Reply 4
Yeah, sorry. :redface:
Reply 5
dvs
Yeah, sorry. :redface:

It's ok, just checking. Thanks. Will rep you when I can.