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10-2. Parabola, Ellipse, Hyperbola
medium
The radius of the circle having its centre at $(0, 3)$ and passing through the foci of the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1$, is
A
$3$
B
$3.5$
C
$4$
D
$\sqrt {12} $
(IIT-1995)
Solution
(c) The co-ordinates of foci are $( \pm ae,\,0)$. Here $a = 4,\,\,b = 3$
$\therefore$ ${b^2} = {a^2}(1 – {e^2})$ ==> $9 = 16(1 – {e^2})$ ==> $\frac{9}{{16}} = 1 – {e^2}$
==> $e = \pm \sqrt {\frac{7}{4}} $; $\therefore$ Points are $( \pm \sqrt 7 ,0)$.
$\therefore$ Radius $ = \sqrt {{{(\sqrt 7 – 0)}^2} + {{(0 – 3)}^2}} = \sqrt {7 + 9} = \sqrt {16} = 4$.
Standard 11
Mathematics
