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Thread: Ellipse and locus of point of intersection of two tangents

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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 $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=0$ if $\displaystyle a^2l^2+b^2m^2=n^2$.
    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 $\displaystyle a^2x^2+b^2y^2=a^4+b^4$

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

    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 $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=0$ if $\displaystyle a^2l^2+b^2m^2=n^2$.
    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 $\displaystyle a^2x^2+b^2y^2=a^4+b^4$

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

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

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