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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Given the points $A(0,4)$ and $B(0,-4)$ the equation of the locus of the point $P(x,y)$ such that $|AP-BP|=6$ is

$\begin{array}{1 1}(A)\;\frac{y^2}{9}+\frac{x^2}{7}=1\\(B)\;\frac{y^2}{7}+\frac{x^2}{9}=1\\(C)\;\frac{x^2}{9}-1\\(D)\;\frac{y^2}{9} -\frac{x^2}{7}=1 \end{array}$

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1 Answer

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=>$AP-BP=\pm 6$
=>$ AP=BP\pm 6$
=> $ \sqrt {x^2+(y-4)^2}=\sqrt {x^2+(y+4)^2} \pm 6$
=> $x^2+y^2-8y+16=x^2+y^2+8y+16+36 \pm 12 \sqrt {x^2+(y+4)^2}$
$-16y-36=\pm 12 \sqrt {x^2+(y+4)^2}$
=>$ 4y+9=\pm 3 \sqrt {x^2+(y+4)^2}$
=> $ 16y^2+72y+81=9(x^2+y^2+8y+16)$
=> $\large\frac{y^2}{9} -\frac{x^2}{7}$$=1$
Hence d is the correct answer.
answered Apr 8, 2014 by meena.p

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