Question

Let \( A(10, 0) \) and \( B(0, \beta) \) 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 \( P \), then the length of the latus rectum of \( E \) is equal to

• A. \( \frac{25}{3} \)
• B. \( \frac{32}{9} \)
• C. \( \frac{32}{5} \)
• D. \( \frac{25}{9} \)

Answer 1

The Correct Option is C

Solution: Substitute \(x = 0\) and \(y = \beta\) in the line equation \(5x + 7y = 50\) to find \(\beta\):

\(7\beta = 50 \Rightarrow \beta = \frac{50}{7}.\)

Thus, \(B = \left(0, \frac{50}{7}\right).\)

Using the section formula, \(P = (3, 5)\), which divides \(AB\) in the ratio \(7 : 3\).

The directrix is \(x = \frac{25}{3}\), so \(a = \frac{25}{3}\) and \(e = \frac{3a}{25}\). Given that \(ae = 3\), solving yields \(a = 5\) and \(b = 4\).

The length of the latus rectum \(LR\) is:

\(LR = \frac{2b^2}{a} = \frac{32}{5}.\)