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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

Find the value of a,b whose foci are (6,4) and (-4,4) and ecentricity is 2?

$\begin{array}{1 1}(a)\;a=\large\frac{6}{7},\normalsize b=\large\frac{4\sqrt{3}}{5}\\(b)\;a=\large\frac{5}{2},\normalsize b=\large\frac{5\sqrt{3}}{2}\\(c)\;a=\large\frac{7}{5},\normalsize b=\large\frac{5\sqrt{7}}{2}\\(d)\;\text{None of these}\end{array}$

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1 Answer

Distance between foci=2ae
$\sqrt{(6+4)^2+(4-4)^2}=2ae$
$10=2\times a\times 2$
$a=\large\frac{5}{2}$
$b^2=a^2(e^2-1)$
$b^2=\large\frac{25}{4}$$(e^2-1)$
$b^2=\large\frac{75}{4}$
$b=\large\frac{5\sqrt 3}{2}$
Hence (b) is the correct answer.
answered Feb 10, 2014 by sreemathi.v
 

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