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  • The radius of the circle passing through the foci of the ellipse <img alt="\frac{x^2}{16}+\frac{y^2}{9}=1" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7Bx%5E2%7D%7B16%7D&

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

Option: 1

\begin{aligned} & 4 \\ \end{aligned}


Option: 2

3 \\


Option: 3

\sqrt{\frac{1}{2}} \\


Option: 4

\frac{7}{2}


Answers (1)

best_answer

For ellipse \frac{x^2}{4^2}+\frac{y^2}{3^2}=1, a=4, b=3

\Rightarrow e=\sqrt{1-\left(\frac{3}{4}\right)^2}=\frac{\sqrt{7}}{4}\\\\ \therefore Foci \: \: are (\sqrt{7}, 0)$ and $(-\sqrt{7}, 0).

 

Center of circle is at (0, 3) and it passes through ,( \pm \sqrt{7}, 0)

therefore radius of circle  =\sqrt{(\sqrt{7})^2+(3)^2}=4

Posted by

avinash.dongre

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