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(A) The given equation of the conic is $9x^{2} + 16y^{2} = 144$.Dividing by $144$,we get $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$.This is the standard

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$8$

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(A) The given equation of the conic is $9x^{2} + 16y^{2} = 144$.Dividing by $144$,we get $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$.This is the standard equation of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,where $a^{2} = 16$ and $b^{2} = 9$.Thus,$a = 4$ and $b = 3$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$.The coordinates of the foci are $(\pm ae, 0) = (\pm 4 	imes \frac{\sqrt{7}}{4}, 0) = (\pm \sqrt{7}, 0)$.Since $A = (\sqrt{7}, 0)$ and $B = (-\sqrt{7}, 0)$ are the foci of the ellipse,by the definition of an ellipse,the sum of the distances from any point $P$ on the ellipse to the two foci is constant and equal to the length of the major axis,which is $2a$.Therefore,$PA + PB = 2a = 2 	imes 4 = 8$.

If the coordinates of two points $A$ and $B$ are $(\sqrt{7}, 0)$ and $(-\sqrt{7}, 0)$ respectively and $P$ is any point on the conic $9x^{2} + 16y^{2} = 144$,then $PA + PB$ is equal to

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