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Thread: Hyperbola and circle

  1. #1
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    Hyperbola and circle

    Hyperbola $\displaystyle x^2-y^2+ax+by=0$ and circle $\displaystyle x^2+y^2=a^2+b^2$ intersect at four points. Show that three of them are vertices of an equilateral triangle.
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    Hyperbola and circle

    Hello atreyyu
    Quote Originally Posted by atreyyu View Post
    Hyperbola $\displaystyle x^2-y^2+ax+by=0$ and circle $\displaystyle x^2+y^2=a^2+b^2$ intersect at four points. Show that three of them are vertices of an equilateral triangle.
    Let $\displaystyle x=r\cos \theta$ and $\displaystyle y=r\sin\theta$, where $\displaystyle r^2=a^2+b^2$.

    Then $\displaystyle (x,y)$ lies on the circle for all $\displaystyle \theta$.

    Circle intersects hyperbola where:

    $\displaystyle r^2\cos^2\theta - r^2\sin^2\theta = -ar\cos\theta-br\sin\theta$

    $\displaystyle \implies \cos^2\theta - \sin^2\theta = -(\frac{a}{r}\cos\theta+\frac{b}{r}\sin\theta)$

    Now express the LHS as $\displaystyle \cos 2\theta$, and the RHS as $\displaystyle -\cos(\theta -\alpha)$, where $\displaystyle \tan\alpha=...$

    Then show that there are three values of $\displaystyle \theta$ that differ by $\displaystyle \frac{2\pi}{3}$ and $\displaystyle \frac{4\pi}{3}$.

    Can you do it now?

    Grandad
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