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ઉપવલય $\frac{{{x^2}}}{{16}}\,\, + \;\,\frac{{{y^2}}}{9}\,\, = \,\,1$ની નાભિઓમાંથી પસાર થતાં અને $(0, 3)$ કેન્દ્ર ધરાવતા વર્તૂળની ત્રિજ્યા =
$4$
$3$
$\sqrt {12} $
$\frac{7}{2}$
Solution
Calculating the radius of the circle:-
Given:-
Ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ and its center $(0,3)$
Find:-
Radius of circle
The equation of the ellipse is $\frac{x^2}{16}+\frac{y^2}{9}=1$
$a=4$ and $b=3$
Eccentricity:-
$e =\sqrt{1-\frac{b^2}{a^2}}$
$=\sqrt{1-\frac{9}{16}}$
$=\sqrt{\frac{7}{16}}$
$=\frac{\sqrt{7}}{4}$
Foci of the ellipse are $(\pm a e, 0)=(\pm \sqrt{7}, 0)$
The radius of the required circle $=$
$\sqrt{(\sqrt{7}-0)^2+(0-3)^2}=\sqrt{7+9}=\sqrt{16}=4$
Hence, The radius of the circle passing through the foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ and having its centre $(0,3)$ is $4$ .
