Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 10 - Parametric Equations and Polar Coordinates - Review - Exercises - Page 711: 54

Answer

$y=mx \pm \sqrt{a^2m^2+b^2}$

Work Step by Step

The standard equation of the ellipse is: $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ ...(I) The equation of the tangent with slope $m$ can be written as: $y=mx+c$ ...(II) From the equation (1) , we have $\dfrac{x^2}{a^2}+\dfrac{(mx+c)^2}{b^2}=1$ $\implies (\dfrac{1}{a^2}+\dfrac{m^2}{b^2})x^2+\dfrac{2mcx}{b^2}+(\dfrac{c^2}{b^2}-1)=0$ Let us find out the discriminant. That is, $D=b^2-4ac=0$ or, $(\dfrac{2mc}{b^2})^2+4(\dfrac{1}{a^2}+\dfrac{m^2}{b^2})(\dfrac{c^2}{b^2}-1)=0$ Then, we have $-(\dfrac{1}{a^2b^2})c^2=-(\dfrac{1}{a^2}-\dfrac{m^2}{b^2})$ $\implies c^2=b^2+a^2m^2$ $\implies c=\pm \sqrt{a^2m^2+b^2``}$ Hence, we have found the equation of the tangent with slope $m$: $y=mx \pm \sqrt{a^2m^2+b^2}$
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