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(B) The given equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$.Here,$a^2=25$ and $b^2=16$,so $a=5$ and $b=4$. The centre $C$ of the ellips

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$9$

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(B) The given equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$.Here,$a^2=25$ and $b^2=16$,so $a=5$ and $b=4$. The centre $C$ of the ellipse is $(0,0)$.The distance $CP$ of a point $P(x, y)$ from the centre $C(0,0)$ is given by $CP = \sqrt{x^2+y^2}$.The maximum distance from the centre to the ellipse occurs at the vertices of the major axis,which are $(\pm 5, 0)$. Thus,the maximum value of $CP$ is $a = 5$.The minimum distance from the centre to the ellipse occurs at the vertices of the minor axis,which are $(0, \pm 4)$. Thus,the minimum value of $CP$ is $b = 4$.The sum of the maximum and minimum values of $CP$ is $a+b = 5+4 = 9$.

If a point $P(x, y)$ moves along the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ and if $C$ is the centre of the ellipse,then the sum of maximum and minimum values of $CP$ is

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