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

Certainly! Here's the solution to the given problem formatted in HTML for CKEditor: ```html

To solve this problem, we need to go through the following steps:

1. Find the coordinates of the point \( B \): The points \( A(10, 0) \) and \( B(0, \beta) \) lie on the line given by the equation: \(5x + 7y = 50\).
   • Substituting \( x = 0 \) to find \( \beta \), we get: \(5(0) + 7\beta = 50 \Rightarrow \beta = \frac{50}{7}\).
   • Thus, the coordinates of \( B \) are \( \left(0, \frac{50}{7}\right) \).
2. Find the coordinates of point \( P \) that divides \( AB \): The point \( P \) divides the line segment \( AB \) in the ratio \( 7:3 \).
   • Using the section formula, the coordinates of \( P \) are: \(\left( \frac{7 \times 0 + 3 \times 10}{7+3}, \frac{7 \times \frac{50}{7} + 3 \times 0}{7+3} \right)\)
   • Calculating individually,
     • \( x \)-coordinate: \(\frac{30}{10} = 3\)
     • \( y \)-coordinate: \(\frac{50}{10} = 5\)
   • Hence, \( P \) is at \( (3, 5) \).
3. Determine the focus \( S \) of the ellipse corresponding to the directrix:
   • The directrix is given by \( 3x - 25 = 0 \), which simplifies to \( x = \frac{25}{3} \).
   • The perpendicular from focus \( S \) on the x-axis passes through \( (3, 5) \).
   • If the focus \( S \) has coordinates \( (c, 0) \), then \( c = 3 \).
4. Find the length of the latus rectum of the ellipse:
   • The relation between focus, directrix, and semi-major axis \( a \) is: \(e = \frac{c}{a} = \frac{25/3 - 3}{a}\Rightarrow a = \frac{9}{2}\).
   • For an ellipse, the length of the latus rectum \( = \frac{2b^2}{a} \) and \(b^2 = a^2(1 - e^2)\) where \( e \) is the eccentricity.
   • Since \( (c, 0) \) is \( (3, 0) \), we deduce \({\|c} = \| a\cdot e\), so \(e = \frac{2}{3}\).
   • Thus, \( b^2 = a^2\left(1 - \frac{4}{9}\right) = \frac{25}{9}\cdot \frac{5}{9} = \frac{25}{81}\cdot 5 \Rightarrow b^2 = \frac{125}{81} \)
   • Therefore, the length of the latus rectum is \(\frac{2b^2}{a}\\) \(= \frac{2\cdot \frac{125}{81}}{\frac{9}{2}} = \frac{32}{5}\).

Therefore, the length of the latus rectum of the ellipse is \(\frac{32}{5}\)

Answer 2

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:

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