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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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If vertex and focus of hyperbola are $(2,3)$ and $(6,3)$ respectively and eccentricity e of the hyperbola is 2 then equation of the hyperbola is

$\begin{array}{1 1}(A)\;\large\frac{(x+1)^2}{16} - \frac{(y-3)^2}{48}=1 \\(B)\;\large\frac{(x+2)^2}{9} - \frac{(y-3)^2}{27}=1 \\(C)\;\large\frac{(x+2)^2}{16} - \frac{(y-3)^2}{48}=1 \\(D)\;none\;of\;these \end{array}$

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1 Answer

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Distance between vertex and focus is $a(e-1)=4$
$=> a=4$
$\therefore b^2=a^2(e^2-1)=48$
Distance of vertex from center is a.
Coordinate of the center is $(-2,3)$
Equation of the required hyperbola
$\large\frac{(x+2)^2}{16} - \frac{(y-3)^2}{48}=1$
Hence C is the correct answer.
answered Apr 21, 2014 by meena.p
 

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