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  • Find the area enclosed between the parabola y^2 = 4ax and the line y = mx.

Q : 6 Find the area enclosed between the parabola  \small y^2=4ax  and the line \small y=mx. 

Answers (1)

best_answer

We have to find the area of the shaded region OBA

The curves y=mx and y2=4ax intersect at the following points

\left ( 0,0 \right )and\left ( \frac{4a}{m^{2}},\frac{4a}{m} \right )

\\y^{2}=4ax\\ \Rightarrow y=2\sqrt{ax}

The required area is

\\\int_{0}^{\frac{4a}{m^{2}}}(2\sqrt{ax}-mx)\\ =2\sqrt{a}[\frac{2x^{\frac{3}{2}}}{3}]_{0}^{\frac{4a}{m^{2}}}-m[\frac{x^{2}}{2}]_{0}^{\frac{4a}{m^{2}}}\\ =\frac{32a^{2}}{3m^{3}}-\frac{8a^{2}}{m^{3}}\\ =\frac{8a^{2}}{3m^{3}}units

Posted by

Sayak

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