Question

The area of a triangle formed by the lines joining the vertex of the parabola \( x^2 = \lambda y \) to the ends of its latus rectum is 18 sq units then the value of \( \lambda \) is

• A. 24
• B. 12
• C. 6
• D. 18

Hint:

For \( x^2 = 4ay \), endpoints of latus rectum are \( (2a,a) \) and \( (-2a,a) \).

Answer 1

The Correct Option is B

Step 1: Compare the given parabola with standard form.
The given parabola is:
\[ x^2 = \lambda y \]
The standard form of parabola is:
\[ x^2 = 4ay \]

Step 2: Find the value of \( a \).
Comparing both equations:
\[ 4a = \lambda \]
\[ a = \frac{\lambda}{4} \]

Step 3: Write the vertex of the parabola.
For \( x^2 = 4ay \), the vertex is:
\[ (0,0) \]

Step 4: Find the endpoints of latus rectum.
For parabola \( x^2 = 4ay \), endpoints of latus rectum are:
\[ (2a,a) \quad \text{and} \quad (-2a,a) \]
Substituting \( a = \frac{\lambda}{4} \):
\[ \left(\frac{\lambda}{2},\frac{\lambda}{4}\right) \quad \text{and} \quad \left(-\frac{\lambda}{2},\frac{\lambda}{4}\right) \]

Step 5: Find the base of the triangle.
The base is the distance between the endpoints of the latus rectum:
\[ \text{Base} = \frac{\lambda}{2} - \left(-\frac{\lambda}{2}\right) \]
\[ \text{Base} = \lambda \]

Step 6: Find the height of the triangle.
The height is the perpendicular distance from vertex \( (0,0) \) to the line \( y = \frac{\lambda}{4} \):
\[ \text{Height} = \frac{\lambda}{4} \]

Step 7: Use area of triangle.
\[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \]
\[ 18 = \frac{1}{2} \times \lambda \times \frac{\lambda}{4} \]
\[ 18 = \frac{\lambda^2}{8} \]
\[ \lambda^2 = 144 \]
\[ \lambda = 12 \]
\[ \boxed{12} \]