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Math Help - Conics Questions Help!!

  1. #1
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    Conics Questions Help!!

    Please help with this question:

    The ellipse x^2/a^2 + y^2/b^2 = 1 cuts the x axis at A and B. There is a vertical tangent to the ellipse at A and also at B. The tangent to the ellipse at the point P (a cos t, b sin t) cuts the tangents through A and B at Q and R respectively. Prove that dAQ.dBR = b^2

    Thanks
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  2. #2
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    Quote Originally Posted by jeta View Post
    Please help with this question:

    The ellipse x^2/a^2 + y^2/b^2 = 1 cuts the x axis at A and B. There is a vertical tangent to the ellipse at A and also at B. The tangent to the ellipse at the point P (a cos t, b sin t) cuts the tangents through A and B at Q and R respectively. Prove that dAQ.dBR = b^2

    Thanks
    One approach would be to get the equation of the tangent at P. You can do this the same way as your hyperbola question: http://www.mathhelpforum.com/math-he...a-problem.html.

    Substitute x = -a into the equation of the tangent to get the y-coord of Q and hence distance AQ.

    Substitute x = a into the equation of the tangent to get the y-coord of R and hence distance BR.

    Now calculate dAQ.dBR and hope you get b^2 as the answer.
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  3. #3
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    Hello,
    Quote Originally Posted by jeta View Post
    Please help with this question:

    The ellipse x^2/a^2 + y^2/b^2 = 1 cuts the x axis at A and B. There is a vertical tangent to the ellipse at A and also at B. The tangent to the ellipse at the point P (a cos t, b sin t) cuts the tangents through A and B at Q and R respectively. Prove that dAQ.dBR = b^2

    Thanks
    I'll let you draw the diagram.

    A and B are on the x axis. This means that A and B have their ordinate equal to 0.
    ---> x/a + 0=1 ---> x=\pm a

    So let \boxed{A(a,0)} and \boxed{B(-a,0)} (the coordinates can be inverted, but I choose them this way. It has no importance)

    If you do an implicit differentiation of the ellipse's equation and that you substitute y=0, you will see that y'=infinity --> vertical

    An equation of a vertical line is in the form x=constant.
    So here, the vertical tangent to the ellipse at A is \boxed{x=a}, since A is on this tangent. Similarly, the vertical tangent to the ellipse at B is \boxed{x=-a}.


    The equation of the tangent at point P is :
    y=y'(a \cos(t))(x-(a \cos(t)))+\underbrace{y(a \cos(t))}_{a \sin(t)}

    By implicit differentiation and rearranging, we have y'(t)=\frac{-b^2 x'(t) x(t)}{a^2 y(t)}
    We know that at point P, x(t)=a \cos(t) and y(t)=b \sin(t). And x'(t)=-a \sin(t)

    Thus y'(a \cos(t))=\frac{b^2a^2 \sin(t) \cos(t)}{a^3 \sin(t)}=\frac{b^2}{a}

    The equation of the tangent at point P is now :
    y=\frac{b^2}{a}(x-a \cos(t))+a \sin(t)=\boxed{\frac{b^2}{a} x-b^2 \cos(t)+a \sin(t)}


    Q is the intersection of this tangent with x=a.
    So y_Q=b^2-b^2 \cos(t)+a \sin(t)
    \boxed{Q(a,b^2-b^2 \cos(t)+a \sin(t))}

    Similarly, x_R=-a and y_R=-b^2-b^2 \cos(t)+a \sin(t)
    \boxed{R(-a,-b^2-b^2 \cos(t)+a \sin(t))}


    Now, find AQ=\sqrt{(x_A-x_Q)^2+(y_A-y_Q)^2} and BR=\dots




    Edit : maybe there's a shorter way, but I decided to do it the straightforward way lol
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  4. #4
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    Thanks alot.
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