Answer :

Given that we need to find the area of the triangle formed by the lines joining the vertex of the parabola x^{2} = 12y to the ends of its latus - rectum.

Comparing it with the standard form of parabola x^{2} = 4by.

⇒ Vertex is 0(0, 0)

⇒ Ends of latus rectum is (2b, b), (- 2b, b)

⇒ 4b = 12

⇒ b = 3

⇒ Ends of latus rectum is (2(3), 3), (- 2(3), 3)

⇒ Ends of latus rectum is A(6, 3), B(- 6, 3)

We know that area of the triangle with the vertices (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) is

⇒

⇒

⇒

⇒

⇒

⇒ A = 18sq.units.

∴The area of the triangle is 18 sq.units.

Rate this question :

How useful is this solution?

We strive to provide quality solutions. Please rate us to serve you better.

Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Expertsview all courses

Dedicated counsellor for each student

24X7 Doubt Resolution

Daily Report Card

Detailed Performance Evaluation

RELATED QUESTIONS :

A parabolic refleRS Aggarwal - Mathematics

A beam is supportRS Aggarwal - Mathematics

The towers of briRS Aggarwal - Mathematics