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Math Help - Hyperbola-locus of P equidist from asymptotes and x-axis?

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    Member ssadi's Avatar
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    Hyperbola-locus of P equidist from asymptotes and x-axis?

    A hyperbola of the form
    \frac{x^2}{\alpha}-\frac{y^2}{\beta}=1
    has asymptotes with equation y^2=m^2 x^2 and passes through the point (a,0). Find an equation of the hyperbola in terms of x, y, a and m.
    Ans:
    \frac{x^2}{a^2}-\frac{y^2}{a^2 m^2}=1
    A point P on this hyperbola is equidistant from one of its asymptoes 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)
    Stuck here please help.
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    Quote Originally Posted by ssadi View Post
    A hyperbola of the form
    \frac{x^2}{\alpha}-\frac{y^2}{\beta}=1
    has asymptotes with equation y^2=m^2 x^2 and passes through the point (a,0). Find an equation of the hyperbola in terms of x, y, a and m.
    Ans:
    \frac{x^2}{a^2}-\frac{y^2}{a^2 m^2}=1
    A point P on this hyperbola is equidistant from one of its asymptoes 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)
    Stuck here please help.
    (Assuming these constans are positive)

    The asymptotes of \frac{x^2}{\alpha} - \frac{y^2}{\beta} = 1 can be obtained by solving \frac{x^2}{\alpha} - \frac{y^2}{\beta} = 0 \implies y^2 = \frac{\beta}{\alpha}x^2.
    Therefore, m^2 = \frac{\beta}{\alpha} \implies \beta = \alpha m^2.
    Now the hyperbola passes through (a,0), therefore \frac{a^2}{\alpha} = 1 \implies \alpha = a^2 \implies \beta = a^2 m^2.
    ---
    Now we do the second half of the problem.

    The P = (x_0,y_0) lie in the first quadrant, and the asympote be y=mx.
    The distance between P and the asymptote is \frac{|y_0 - mx_0|}{\sqrt{1+m^2}}.
    The distance between P and the x-axis is x_0.
    Thus, we want, \frac{|y_0 - mx_0|}{\sqrt{1+m^2}} = x_0 \implies (y_0 - mx_0)^2 = x_0^2 (1+m^2).
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