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Math Help - Ellipse and locus of point of intersection of two tangents

  1. #1
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    Ellipse and locus of point of intersection of two tangents

    Prove that the line lx+my+n=0 is a tangent to the ellipse \frac{x^2}{a^2}+\frac{y^2}{b^2}=0 if a^2l^2+b^2m^2=n^2.
    Tangents are drawn from a point P to the ellipse. Lines from the origin are drawn perpendicular to the tangents. If these lines are conjugate diameters, prove that the equation of the locus of P is a^2x^2+b^2y^2=a^4+b^4

    I have proved the first part.
    For this next part the equation of a tangent to an ellipse in the form
    y=mx\pm\sqrt{a^2m^2+b^2}
    gradients are related by
    \frac{1}{m_1}.\frac{1}{m_2}=-\frac{b^2}{a^2}
    m_2=-\frac{a^2}{b^2{m_1}^2}

    I used the two gradients, m, in two equations for tangents and tried to solve for x and y, am I doing it right? because I get very very long and difficult expressions.
    Thanks!
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  2. #2
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    Quote Originally Posted by arze View Post
    Prove that the line lx+my+n=0 is a tangent to the ellipse \frac{x^2}{a^2}+\frac{y^2}{b^2}=0 if a^2l^2+b^2m^2=n^2.
    Tangents are drawn from a point P to the ellipse. Lines from the origin are drawn perpendicular to the tangents. If these lines are conjugate diameters, prove that the equation of the locus of P is a^2x^2+b^2y^2=a^4+b^4

    I have proved the first part.
    For this next part the equation of a tangent to an ellipse in the form
    y=mx\pm\sqrt{a^2m^2+b^2}
    gradients are related by
    \frac{1}{m_1}.\frac{1}{m_2}=-\frac{b^2}{a^2}
    m_2=-\frac{a^2}{b^2{m_1}^2}
    For the last part, let P be the point (p,q). The tangent lx+my+n=0 satisfies a^2l^2+b^2m^2=n^2, and it passes through P if ap+bq=n. Substitute n from the second of those two equations into the first equation: a^2l^2+b^2m^2 = (lp+mq)^2, which can be written as

    (a^2-p^2)\bigl(\tfrac lm\bigr)^2 -2pq\tfrac lm + (b^2-q^2) = 0.\qquad(*)

    The two roots of the quadratic equation (*) in \tfrac lm are \tfrac lm = -m_1 and \tfrac lm = -m_2, where m_1,\,m_2 are the gradients of the tangents through P (because -\tfrac lm is the gradient of the line lx+my+n=0). Therefore m_1m_2 = \frac{b^2-q^2}{a^2-p^2} (formula for product of roots of quadratic: constant term divided by coefficient of x^2). But you already know that \frac{1}{m_1}\mathord{\cdot}\frac{1}{m_2}=-\frac{b^2}{a^2}. Put those two expressions for m_1m_2 together to see that \frac{b^2-q^2}{a^2-p^2} = -\frac{a^2}{b^2}, from which a^2p^2 + b^2q^2 = a^4+b^4. Finally, replace (p,q) by (x,y) to get the locus of P.
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