Question

If the points of intersection of the parabola y² = 5x and x² = 5y lie on the line L, then the area of the triangle formed by the directrix of one parabola, latus rectum of another parabola and the line L is 

• A. \(\frac{15}{32}\)
• B. \(\frac{12}{25}\)
• C. \(\frac{25}{8}\)
• D. \(\frac{25}{32}\)

Answer 1

The Correct Option is C

To find the area of the triangle formed by the line \( y = x \), the directrix of the parabola \( y^2 = 5x \), and the latus rectum of the parabola \( x^2 = 5y \), we proceed as follows:

1. Intersection Points of the Parabolas:
The parabolas are \( y^2 = 5x \) and \( x^2 = 5y \). Substitute \( x = \frac{y^2}{5} \) into \( x^2 = 5y \):
\( \left( \frac{y^2}{5} \right)^2 = 5y \)
\( \frac{y^4}{25} = 5y \)
\( y^4 = 125y \)
\( y (y^3 - 125) = 0 \)
\( y = 0 \) or \( y = 5 \). For \( y = 0 \), \( x = 0 \); for \( y = 5 \), \( x = \frac{5^2}{5} = 5 \). Thus, intersection points are \((0, 0)\) and \((5, 5)\).

2. Equation of Line \( L \):
The line through \((0, 0)\) and \((5, 5)\) has slope \(\frac{5 - 0}{5 - 0} = 1\), so:
\( y = x \)

3. Directrix of \( y^2 = 5x \):
Rewrite: \( y^2 = 4 \cdot \frac{5}{4} x \), where \( 4a = 5 \), so \( a = \frac{5}{4} \). 
\( x = -a = -\frac{5}{4} \)

4. Latus Rectum of \( x^2 = 5y \):
Rewrite: \( x^2 = 4 \cdot \frac{5}{4} y \), where \( 4a = 5 \), so \( a = \frac{5}{4} \). 
\( y = a = \frac{5}{4} \)

5. Vertices of the Triangle:
Find intersections of \( y = x \), \( x = -\frac{5}{4} \), and \( y = \frac{5}{4} \):

- \( y = x \) and \( x = -\frac{5}{4} \): \( y = -\frac{5}{4} \), so point \( A\left( -\frac{5}{4}, -\frac{5}{4} \right) \).
- \( y = x \) and \( y = \frac{5}{4} \): \( x = \frac{5}{4} \), so point \( B\left( \frac{5}{4}, \frac{5}{4} \right) \).
- \( x = -\frac{5}{4} \) and \( y = \frac{5}{4} \): point \( C\left( -\frac{5}{4}, \frac{5}{4} \right) \).

Vertices are \( A\left( -\frac{5}{4}, -\frac{5}{4} \right) \), \( B\left( \frac{5}{4}, \frac{5}{4} \right) \), \( C\left( -\frac{5}{4}, \frac{5}{4} \right) \).

6. Area of the Triangle:
Use the area formula for vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\):

\( \text{Area} = \frac{1}{2} \left| x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2) \right| \)
Substitute \( A\left( -\frac{5}{4}, -\frac{5}{4} \right) \), \( B\left( \frac{5}{4}, \frac{5}{4} \right) \), \( C\left( -\frac{5}{4}, \frac{5}{4} \right) \):

\( y_2 - y_3 = \frac{5}{4} - \frac{5}{4} = 0 \)
\( y_3 - y_1 = \frac{5}{4} - \left( -\frac{5}{4} \right) = \frac{10}{4} = \frac{5}{2} \)
\( y_1 - y_2 = -\frac{5}{4} - \frac{5}{4} = -\frac{10}{4} = -\frac{5}{2} \)
\( \text{Area} = \frac{1}{2} \left| \left( -\frac{5}{4} \right) (0) + \frac{5}{4} \cdot \frac{5}{2} + \left( -\frac{5}{4} \right) \cdot \left( -\frac{5}{2} \right) \right| = \frac{1}{2} \left| 0 + \frac{25}{8} + \frac{25}{8} \right| = \frac{1}{2} \cdot \frac{50}{8} = \frac{25}{8} \)

Final Answer:
The area of the triangle is \(\frac{25}{8}\).