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    given that S and S' are the focii of hyp with eq/n x^2/a^2 - y^2/b^2 = 1 show that SPand S'P are eqaully inclined to the tangent at any point P on the curve

    another q

    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

    thanks alot :secruity:

    Going to bring this back to the top, as also would like help on this question
    thankyou , Ilana

    For the 1st question, you can use the reflection propety of a hyperbola.

    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.

    If you can't use the reflection property, here's a solution to the problem.
    Attached Images
  1. File Type: pdf Hyperbola - Reflection Property.pdf (87.0 KB, 155 views)
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Updated: February 17, 2005

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