p5 hyperbolas help pleaseWatch
prove that the triangle formed formed by the asymptotes of the curve with eq/n x^2 - 2y^2 = 4 and any tangent to the curve is of constant area
thankyou , Ilana
Can you just quote it? Otherwise you may have to go through a lot of trig.
Let P be a point on the curve with coords (x0, y0). Then the equation of the tangent at P is:
x(x0) - 2y(y0) = 4
And the equation of the asymptotes is:
x^2 = 2y^2
Let the asymptotes intersect the tangent at point P, then:
(x^2)/2 = y^2
x(x0) - 4 = 2y(y0)
x(x0) - 4 = 2(sqrt[x^2/2])(y0)
[x(x0) - 4]^2 = 4x^2(y0)/2 = 2x^2(y0)^2
2x^2(y0)^2 - x^2(x0)^2 + 8x(x0) - 16 = 0
x^2[2(y0)^2 - (x0)^2] + 8x(x0) - 16 =0
If you compare 2(y0)^2 - (x0)^2 to the equation of the curve, you'll see that 2(y0)^2 - (x0)^2=-4. So:
-4x^2 + 8(x0)x - 16 =0
x^2 - 2(x0)x + 4 = 0
This equation has roots a & b, which are the coordinates of the point of intersection between the tangent and the asymptotes. a*b=4, so the product of the x-coordinates is always constant, so the lengths of the lines connecting the origin to the vertices of the triangle is constant. So the triangle is of constant area.
Hopefully my reasoning is too off.