If the equation of the hyperbola having (8,3) | vedclass

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(C) The foci are $F_1(8,3)$ and $F_2(0,3)$. The center of the hyperbola is the midpoint of the foci: $(\frac{8+0}{2}, \frac{3+3}{2}) = (4,3)$. Thus,$\

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$\alpha^2$

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(C) The foci are $F_1(8,3)$ and $F_2(0,3)$. The center of the hyperbola is the midpoint of the foci: $(\frac{8+0}{2}, \frac{3+3}{2}) = (4,3)$. Thus,$\alpha = 4$ and $\beta = 3$.The distance between the foci is $2ae = 8 - 0 = 8$. Given $e = \frac{4}{3}$,we have $2a(\frac{4}{3}) = 8$,which implies $a = 3$.Using the relation $b^2 = a^2(e^2 - 1)$,we get $b^2 = 3^2((\frac{4}{3})^2 - 1) = 9(\frac{16}{9} - 1) = 9(\frac{7}{9}) = 7$.In the standard form $\frac{(x-\alpha)^2}{a^2} - \frac{(y-\beta)^2}{b^2} = 1$,we have $p = a^2 = 9$ and $q = b^2 = 7$.Therefore,$p+q = 9+7 = 16$.Since $\alpha = 4$ and $\beta = 3$,$\alpha^2 = 16$. Thus,$p+q = \alpha^2$.

If the equation of the hyperbola having $(8,3)$ and $(0,3)$ as foci and $\frac{4}{3}$ as eccentricity is $\frac{(x-\alpha)^2}{p}-\frac{(y-\beta)^2}{q}=1$,then $p+q=$

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