Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter P - Problem Solving - Page 40: 15

Answer

Please see below.

Work Step by Step

As we know, the distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. So, we can calculate $d_1$ and $d_2$ as follows.$$d_1=\sqrt{(x-(-1))^2+(y-0)^2}=\sqrt{(x+1)^2+y^2} \\ d_2= \sqrt{(x-1)^2+(y-0)^2}= \sqrt{(x-1)^2+y^2}$$By applying the hypothesis $d_1d_2=1$, we conclude that$$d_1^2d_2^2=1 \quad \Rightarrow \quad \left ( (x+1)^2+y^2 \right ) \left ( (x-1)^2+y^2 \right )=x^4+x^2-2x^3+x^2y^2+x^2-2x+1+y^2+2x^3+2x-4x^2+2xy^2+y^2x^2+y^2-2xy^2+y^4=1 \quad \Rightarrow \quad x^4 +2x^2y^2+y^4-2x^2+2y^2=0 \quad \Rightarrow \quad (x^2+y^2)^2=2(x^2-y^2).$$
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