Let A(alpha, 0) and B(0, beta) be the points | vedclass

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(D) The line is $5x + 7y = 50$. For $A(\alpha, 0)$,$5\alpha = 50 \implies \alpha = 10$,so $A = (10, 0)$. For $B(0, \beta)$,$7\beta = 50 \implies \beta

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$\frac{32}{5}$

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(D) The line is $5x + 7y = 50$. For $A(\alpha, 0)$,$5\alpha = 50 \implies \alpha = 10$,so $A = (10, 0)$. For $B(0, \beta)$,$7\beta = 50 \implies \beta = \frac{50}{7}$,so $B = (0, \frac{50}{7})$.Point $P$ divides $AB$ in the ratio $7:3$ internally. Using the section formula,$P = \left( \frac{7(0) + 3(10)}{7+3}, \frac{7(\frac{50}{7}) + 3(0)}{7+3} ight) = \left( \frac{30}{10}, \frac{50}{10} ight) = (3, 5)$.The directrix is $x = \frac{25}{3}$. The focus $S$ lies on the $x$-axis,so $S = (ae, 0)$. The perpendicular from $S$ to the $x$-axis is the line $x = ae$. Since this passes through $P(3, 5)$,we have $ae = 3$.The equation of the directrix is $x = \frac{a}{e} = \frac{25}{3}$.From $ae = 3$ and $\frac{a}{e} = \frac{25}{3}$,we multiply them: $a^2 = 3 	imes \frac{25}{3} = 25 \implies a = 5$.Then $e = \frac{3}{a} = \frac{3}{5}$.Using $b^2 = a^2(1 - e^2)$,we get $b^2 = 25(1 - \frac{9}{25}) = 25(\frac{16}{25}) = 16$,so $b = 4$.The length of the latus rectum is $\frac{2b^2}{a} = \frac{2(16)}{5} = \frac{32}{5}$.

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