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Math Help - Conics Q

  1. #1
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    Conics Q

    The tangent at P(a\sec\theta,b\tan\theta) on the hyperbola \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 has equation \frac{x\sec\theta}{a}-\frac{y\tan\theta}{b}=1.

    If teh tangent at P is also tangent to the circle center (ae,0) and radius ae\sqrt{e^2+1}, show that \sec\theta=-e where e is the eccentricity of the hyperbola.
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  2. #2
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    Hello vuze88
    Quote Originally Posted by vuze88 View Post
    The tangent at P(a\sec\theta,b\tan\theta) on the hyperbola \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 has equation \frac{x\sec\theta}{a}-\frac{y\tan\theta}{b}=1.

    If teh tangent at P is also tangent to the circle center (ae,0) and radius ae\sqrt{e^2+1}, show that \sec\theta=-e where e is the eccentricity of the hyperbola.
    Are you sure you haven't made a mistake in the question? I make it that the radius of the circle should be a\sqrt{e^2+1}.

    You then solve the problem by saying that a line is a tangent to a circle if and only if the distance from the centre of the circle to the line is equal to the radius of the circle. So in this case we get (using the distance formula \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}} with the point (ae,0) and the line \frac{x\sec\theta}{a}-\frac{y\tan\theta}{b}=1):

    \frac{|e\sec\theta-1|}{\sqrt{\frac{\sec^2\theta}{a^2}+\frac{\tan^2\th  eta}{b^2}}}=a\sqrt{e^2+1}

    \Rightarrow\frac{a^2b^2(e\sec\theta-1)^2}{b^2\sec^2\theta+a^2\tan^2\theta}=a^2(e^2+1)

    \Rightarrow\frac{a^2(e^2-1)(e\sec\theta-1)^2}{a^2(e^2-1)\sec^2\theta+a^2(\sec^2\theta-1)}=e^2+1, using b^2 = a^2(e^2-1) and \tan^2\theta=\sec^2\theta -1

    \Rightarrow\frac{(e^2-1)(e\sec\theta-1)^2}{e^2\sec^2\theta-1}=e^2+1

    \Rightarrow\frac{(e^2-1)(e\sec\theta-1)^2}{(e\sec\theta-1)(e\sec\theta+1)}=e^2+1

    \Rightarrow\frac{(e^2-1)(e\sec\theta-1)}{(e\sec\theta+1)}=e^2+1

    \Rightarrow(e^2-1)(e\sec\theta-1)=(e\sec\theta+1)(e^2+1)

    \Rightarrow e\sec\theta([e^2-1]-[e^2+1])=[e^2+1]+[e^2-1]

    \Rightarrow -2e\sec\theta = 2e^2

    \Rightarrow \sec\theta=-e

    Grandad
    Last edited by mr fantastic; September 8th 2009 at 01:39 AM. Reason: No edit - just flagging the post as having been moved from another thread.
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  3. #3
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    Quote Originally Posted by vuze88 View Post
    The tangent at P(a\sec\theta,b\tan\theta) on the hyperbola \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 has equation \frac{x\sec\theta}{a}-\frac{y\tan\theta}{b}=1.

    If the tangent at P is also tangent to the circle center (ae,0) and radius ae\sqrt{e^2+1}, show that \sec\theta=-e where e is the eccentricity of the hyperbola.
    The condition for the line to be tangent to the circle is that the distance from the centre of the circle to the line should be equal to the radius of the circle. But the formula for the distance d from the point (x_0,y_0) to the line px+qy+r=0 is d = \frac{|px_0+qy_0+r|}{\sqrt{p^2+q^2}}.

    So the distance from (ae,0) to the line \frac{x\sec\theta}{a}-\frac{y\tan\theta}{b}-1 = 0 is |e\sec\theta-1|\above1pt\displaystyle\sqrt{\frac{\sec^2\theta}{  a^2} + \frac{\tan^2\theta}{b^2}}. We want this to be equal to ae\sqrt{e^2+1}. That gives the equation

    |e\sec\theta-1| = ae\sqrt{e^2+1}\sqrt{\frac{b^2\sec^2\theta+a^2\tan^  2\theta}{a^2b^2}}.

    Simplify the big square root on the right-hand side of that equation as much as you can, using the relations b^2 = a^2(e^2 - 1) and \sec^2\theta-\tan^2\theta=1. Then square both sides of the equation, and solve it for \sec\theta. You should find that there are two solutions, \sec\theta = 1/e and \sec\theta=-e. But the first of those solutions cannot occur, because |\sec\theta|\geqslant1 and 1/e<1.
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  4. #4
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    thanks for the help and sorry about the mistake
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  5. #5
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    thanks alot
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