Using integration, find the area of the parabola y2 = 4ax bounded by its latus rectum.

 

 
 
 
 
 

Answers (1)
S safeer

Parabola: y^2= 4ax....(i)

Latus ractum: x=a

Intersection Point 

Put x =a in equation (i)

\\ y^2 = 4a^2 \\ y = \pm 2a

Intersection Point A(a,2a), B(a,-2a)

\\ $Required Area $ = 2 $Area of AOM$ \\ \\ = 2\int _0^{a} \sqrt{4ax}dx \\\\ = 2 \times \frac{2}{3 \times 4a} \left[(4ax)^\frac{3}{2}\right]_0^{a}\\\\ = \frac{8}{3}a^2

 

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