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

The radius of the circle passing through the foci of the ellipse $\large\frac{x^2}{16}+\frac{y^2}{9}$$=1$ and having its center at $(0,3)$ is

$\begin{array}{1 1}(A)\;4\\(B)\;3 \\(C)\;\sqrt {\large\frac{1}{2}} \\(D)\;\large\frac{7}{2} \end{array}$

1 Answer

For ellipse $\large\frac{x^2}{4^2}+\frac{y^2}{3^2}$$=1$
=>$e =\sqrt {1-\bigg(\large\frac{3}{4} \bigg)^2}$
$\qquad= \sqrt {\large\frac{7}{4}}$
Foci are $(\sqrt 7,0)$ and $(-\sqrt 7,0)$
Centre of circle is at $(0,3)$ and it passes through radius of circle
$\qquad= \sqrt{(7)^2+(3)^2}$
$\qquad= 4$
Hence A is the correct answer.
answered Apr 9, 2014 by meena.p

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