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Math Help - [SOLVED] Hyperbola

  1. #1
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    [SOLVED] Hyperbola

    Hello.

    For this question - Show how to determine a point P of the hyperbola xy = c^2 such that the asymptotes cut off a segment of length 6c in the tangent at P,

    I've equated 6c^2= sqrt[a^2+b^2], with the x intercept being (a,0) and the y intercept being (0,b) BUT that gets me nowhere cause I now have 3 variables to work with.

    I tried finding the equation of the tangent, which is y = b/a * x + b.

    Please could someone show me how to do this?
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  2. #2
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    Hello UltraGirl
    Quote Originally Posted by UltraGirl View Post
    Hello.

    For this question - Show how to determine a point P of the hyperbola xy = c^2 such that the asymptotes cut off a segment of length 6c in the tangent at P,

    I've equated 6c^2= sqrt[a^2+b^2], with the x intercept being (a,0) and the y intercept being (0,b) BUT that gets me nowhere cause I now have 3 variables to work with.

    I tried finding the equation of the tangent, which is y = b/a * x + b.

    Please could someone show me how to do this?
    I assume that you know how to use the parametric equations of the rectangular hyperbola:
    x = ct, y = \frac ct
    From these we get:
    \frac{dx}{dt}= c, \frac{dy}{dt} = -\frac c{t^2}

    \Rightarrow \frac{dy}{dx}= \frac{dy}{dt}\div\frac{dx}{dt}
    = -\frac1{t^2}
    So the equation of the tangent at \Big(ct, \frac ct\Big) is:
    y - \frac ct = -\frac1{t^2}\Big(x - ct\Big)
    i.e.
    t^2y+x = 2ct
    The intercepts of this line with the coordinate axes are \Big(0,\frac{2c}{t}\Big) and \Big(2ct,0\Big). And if the length of this line segment is 6c, then:
    (6c)^2 = \Big(\frac{2c}{t}\Big)^2+(2ct)^2

    \Rightarrow 36c^2 = \frac{4c^2}{t^2}+4c^2t^2

    \Rightarrow 9t^2 = 1 +t^4

    \Rightarrow (t^2)^2-9t^2+1=0

    \Rightarrow t^2 = \frac{9\pm\sqrt{81-4}}{2}

    \Rightarrow t = \pm\sqrt{\frac{9\pm\sqrt{77}}{2}}
    Thus we have found four values of t, which give the four required points when substituted into the parametric co-ordinates \Big(ct, \frac ct\Big).

    Grandad
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