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Math Help - hyperbola help!

  1. #1
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    hyperbola help!

    P(a\sec\theta, b\tan\theta) is a point on the hyperbola \frac{x^2}{a^2}-\frac{y^2}{b^2}=1. The ordinate at P and the tangent at P meet the asymptote at Q and R respectively. if the normal at P meets the x-axis at N prove that RQ and QN are perpendicular.

    The ordinate at P is its y-value.
    Asymptotes: y=\pm\frac{b}{a}x
    when y=b\tan\theta, x=\frac{a}{b}y=a\tan\theta
    Q(a\tan\theta,b\tan\theta)
    equation of tangent bx\sec\theta-ay\tan\theta=ab
    when y=\frac{b}{a}x
    bx(\sec\theta-\tan\theta)=ab
    x=\frac{a}{\sec\theta-\tan\theta}
    y=\frac{b}{a}(\frac{a}{\sec\theta-\tan\theta})=\frac{b}{\sec\theta-\tan\theta}
    R(\frac{a}{\sec\theta-\tan\theta},\frac{b}{\sec\theta-\tan\theta})
    equation of normal by\sec\theta+ax\tan\theta=(a^2+b^2)\tan\theta\sec\  theta
    when y=0, x=\frac{a^2+b^2}{a}\sec\theta
    N(\frac{a^2+b^2}{a}\sec\theta,0)

    gradient RQ =\frac{b}{a}
    gradient QN =\frac{b\tan\theta}{a\tan\theta-\frac{a^2+b^2}{a}\sec\theta}
    my problem is i can't get the product of the two gradients to be -1.
    thanks!
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  2. #2
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    You're good for \theta = \frac{\pi}{2}. Isn't that the point?
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  3. #3
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    In the problem, it is stated that the ordinate at P and the tangent at P meet the asymptote at Q and R respectively.
    I presume that the point Q lies on the asymptote y = bx/a, and R lies on the asymptote y = -bx/a.
    The x co-ordinate of Q is the same as the x co-ordinate of P. Only ordinate will change. So the slope of RQ is not equal to b/a.
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  4. #4
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    Quote Originally Posted by sa-ri-ga-ma View Post
    In the problem, it is stated that the ordinate at P and the tangent at P meet the asymptote at Q and R respectively.
    I presume that the point Q lies on the asymptote y = bx/a, and R lies on the asymptote y = -bx/a.
    The x co-ordinate of Q is the same as the x co-ordinate of P. Only ordinate will change. So the slope of RQ is not equal to b/a.
    If R intersects y=-\frac{b}{a}x, then R(\frac{a}{\sec\theta+\tan\theta},-\frac{b}{\sec\theta+\tan\theta}) which does not help.
    I got the gradient QR
    \frac{b(\sec^2\theta+\sec\theta)}{a(\sec^2\theta+\  sec\theta-2)}
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  5. #5
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    Your co-ordinates of Q are wrong.
    x co-ordinate of Q is a*secθ.
    Equation of asymptote is y = b/a*x
    So y co-ordinate of Q is b/a*a*secθ = b*secθ
    Q = ( a*secθ, b*secθ)
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