Question

Let A(α,0) and B(0,β) be the points on the line 5x + 7y = 50. Let the point P divide the line segment AB internally in the ratio 7: 3. Let 3x - 25 = 0 be a directrix of the ellipse

E: $\frac{x^2}{a^2}$ + $\frac{y^2}{b^2}$ = 1 and the corresponding focus be S. If from S, the perpendicular on the x-axis passes through P, find the length of the latus rectum of E.

Answer 1

$\frac{32}{5}$

Answer 2

1. Find points A and B by substituting y=0 and x=0 into $5x + 7y = 50$. This gives $A(10,0)$ and $B(0, \frac{50}{7})$.
2. Calculate point P using the section formula for internal division in the ratio 7:3: $P = (\frac{7 \cdot 0 + 3 \cdot 10}{7+3}, \frac{7 \cdot \frac{50}{7} + 3 \cdot 0}{7+3}) = (3,5)$.
3. The directrix is $x = \frac{25}{3}$. For an ellipse, the directrix is $x = \frac{a}{e}$, so $\frac{a}{e} = \frac{25}{3}$. The corresponding focus is $S(ae, 0)$.
4. The condition that the perpendicular from S to the x-axis passes through P means $x_S = x_P$, so $ae = 3$.
5. Solve the system: $\frac{a}{e} = \frac{25}{3}$ and $ae = 3$. Multiplying gives $a^2 = 25$, so $a=5$. Substituting $a=5$ into $ae=3$ gives $e=\frac{3}{5}$.
6. Calculate $b^2 = a^2(1-e^2) = 5^2(1 - (\frac{3}{5})^2) = 25(1 - \frac{9}{25}) = 16$.
7. The length of the latus rectum is $LR = \frac{2b^2}{a} = \frac{2 \cdot 16}{5} = \frac{32}{5}$.