Ellipse Ques 9
- The equation of the circle passing through the foci of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ and having centre at $(0,3)$ is
(2013 Main)
(a) $x^{2}+y^{2}-6 y-7=0$
(c) $x^{2}+y^{2}-6 y-5=0$
(b) $x^{2}+y^{2}-6 y+7=0$
(d) $x^{2}+y^{2}-6 y+5=0$
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Answer:
Correct Answer: 9.(a)
Solution:
Formula:
- Given equation of ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$
Here, $a=4, b=3, e=\sqrt{1-\frac{9}{16}} \Rightarrow \frac{\sqrt{7}}{4}$
$\therefore$ Foci $=( \pm a e, 0)= \pm 4 \times \frac{\sqrt{7}}{4}, 0=( \pm \sqrt{7}, 0)$
Radius of the circle, $r=\sqrt{(a e)^{2}+b^{2}}$
$ =\sqrt{7+9}=\sqrt{16}=4 $
Now, equation of circle is
$ \begin{aligned} (x-0)^{2}+(y-3)^{2} & =16 \\ \therefore \quad x^{2}+y^{2}-6 y-7 & =0 \end{aligned} $
BETA
