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: 50

Answer

$(y-1)^2=12(x+1)$

Work Step by Step

Given: Directrix $x=-4$, Focus:$F(2,1)$ Let us consider the standard form for the parabola $(y-k)^2=4p(x-h)$ ; $x=-a$ with Vertices:$(h,k)$ Then, we have $|x+4|=\sqrt{(x-2)^2+(y-1)^2}$ $\implies (x+4)^2=(x-2)^2+(y-1)^2$ This implies that $(y-1)^2=x^2+16+8x-x^2-4+4x$ and $(y-1)^2=12x+12$ Hence, $(y-1)^2=12(x+1)$
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