ઉપવલય frac{{{x^2}}}{{16}},, + ;,frac{{{y^2}}} | vedclass

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Question

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

Vedclass · Accepted Answer

$4$

Vedclass · Answer

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 $4$ .

ઉપવલય $\frac{{{x^2}}}{{16}}\,\, + \;\,\frac{{{y^2}}}{9}\,\, = \,\,1$ની નાભિઓમાંથી પસાર થતાં અને $(0, 3)$ કેન્દ્ર ધરાવતા વર્તૂળની ત્રિજ્યા =

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