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(C) For a parabola $y^2 = 4ax$,the focal distance of a point $(x, y)$ is given by $x + a$. Given $y^2 = 32x$,we have $4a = 32$,so $a = 8$. The focal d

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$272$

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(C) For a parabola $y^2 = 4ax$,the focal distance of a point $(x, y)$ is given by $x + a$. Given $y^2 = 32x$,we have $4a = 32$,so $a = 8$. The focal distance is $x + 8 = 10$,which implies $x = 2$. Substituting $x = 2$ into the parabola equation: $y^2 = 32(2) = 64$,so $y = \pm 8$. Thus,the points are $(2, 8)$ and $(2, -8)$. Here,$x_1 = 2, y_1 = 8$ and $x_2 = 2, y_2 = -8$. We need to calculate $2(x_1^2 + x_2^2 + y_1^2 + y_2^2)$. $x_1^2 + x_2^2 + y_1^2 + y_2^2 = 2^2 + 2^2 + 8^2 + (-8)^2 = 4 + 4 + 64 + 64 = 136$. Therefore,$2(136) = 272$.

If $(x_1, y_1)$ and $(x_2, y_2)$ are the points on the parabola $y^2 = 32x$ each at a focal distance of $10$ units,then $2(x_1^2 + x_2^2 + y_1^2 + y_2^2) = $

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