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

Consider the parabolas:

\[ y^2 = 5x, \quad x^2 = 5y \]

1. Find intersection points:

\[ x = \frac{y^2}{5} \]

Substitute into \( x^2 = 5y \):

\[ \left(\frac{y^2}{5}\right)^2 = 5y \Rightarrow \frac{y^4}{25} = 5y \Rightarrow y^4 = 125y \Rightarrow y(y^3 - 125) = 0 \]

\[ y = 0 \ \text{or} \ y = 5 \]

Corresponding points:

\[ (0,0), \ (5,5) \]

2. Directrix of \( y^2 = 5x \):

\[ x = -\frac{5}{4} \]

3. Latus rectum of \( x^2 = 5y \):

\[ y = \frac{5}{4} \]

4. Triangle is formed by lines:

\[ x = -\frac{5}{4}, \quad y = \frac{5}{4}, \quad y = x \]

Vertices:

\[ \left(-\frac{5}{4}, -\frac{5}{4}\right), \quad \left(-\frac{5}{4}, \frac{5}{4}\right), \quad \left(\frac{5}{4}, \frac{5}{4}\right) \]

5. Area:

\[ \text{Area} = \frac{1}{2} \times \frac{5}{2} \times \frac{5}{2} = \frac{25}{8} \]

Thus, the area is \( \frac{25}{8} \).

Answer 2

To solve this problem, we need to determine the area of the triangle formed by the directrix of one parabola, the latus rectum of another parabola, and the line on which the intersection points of the parabolas lie.

Firstly, consider the given parabolas:

1. The equation for the first parabola is \( y^2 = 5x \). This is a standard form of a parabola that opens to the right.

2. The equation for the second parabola is \( x^2 = 5y \). This is a standard form of a parabola that opens upwards.

Next, we find the points of intersection of the two parabolas. Using

\( x = \frac{y^2}{5} \), \quad \( y = \frac{x^2}{5} \)

Substitute into the equation for \( y \):

\( y = \frac{(y^2/5)^2}{5} = \frac{y^4}{125} \)

This gives \( 125y = y^4 \Rightarrow y(y^3 - 125) = 0 \).

So, \( y = 0 \) or \( y = 5 \).

Using \( x = \frac{y^2}{5} \), we get intersection points: \( (0,0) \) and \( (5,5) \).

The directrix of the parabola \( y^2 = 5x \) is \( x = -\frac{5}{4} \).

The latus rectum of the parabola \( x^2 = 5y \) is \( y = \frac{5}{4} \).

The line through the points \( (0,0) \) and \( (5,5) \) is \( y = x \).

Now find the area of the triangle formed by \( x = -\frac{5}{4} \), \( y = \frac{5}{4} \), and \( y = x \).

The vertices are \( \left(-\frac{5}{4}, -\frac{5}{4}\right) \), \( \left(-\frac{5}{4}, \frac{5}{4}\right) \), and \( \left(\frac{5}{4}, \frac{5}{4}\right) \).

Area:

\[ A = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| \]

\[ A = \frac{1}{2} \left| -\frac{5}{4}\left(\frac{5}{4} - \frac{5}{4}\right) + \left(-\frac{5}{4}\right)\left(\frac{5}{4} + \frac{5}{4}\right) + \frac{5}{4}\left(-\frac{5}{4} - \frac{5}{4}\right) \right| \]

\[ A = \frac{1}{2} \cdot \frac{25}{4} = \frac{25}{8} \]

Thus, the correct answer is \( \frac{25}{8} \).