area of triangle formed by pair of lines (x+2a)^2-3y^2 and x=a

The pair of lines (x+2a)² – 3y² = 0 passes through the origin ( just substitiute 0,0 and if it satisfies then it passes through the origin ). Now this pair of lines along with x=a forms a triangle. x=a must be a line parallel to the x-axis either in the left or right side. The pair of lines is in the form of an “X” with its pt. of intersection at the origin. So, (0,0) is one of the vertex. x=a and (x+2a)² – 3y² = 0 intersect at 2 points.

Solving botht the equations:

Substitute x=a in (x+2a)² – 3y² = 0 as on the pt. of intersection, that pt. should satisfy botht the lines.

So you will get,

(3a)² – 3y² = 0.

So, 3a² = y²

So y = a√3, -a√3

So the points of intersection are (a,a√3) and (a, -a√3).

So now we know all the 3 vertices of the triangle : (0,0), (a,a√3) and (a, -a√3).

 

Now area of the triangle is : 1/2[x₁(y₂-y₃) + x₂(y₃-y₁) + x3(y₁-y₂)]

Substitute the vertices in this equation and find the area. The final answer is 3a²