Find the area enclosed between the parabola y 2 = 4ax and the line y = mx
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The point of intersection on both curves is (0,0) and (4a/m², 4a/m)
A perpendicular AC will be drawn to x-axis to make area OABO= area OCABO - area of ΔOCA
applying limit to the equation of =2√ax . dx- mx dx
= 2√a [ x³/₂/ 3/2]⁴ᵃ/m² - m[x²/2]⁴ᵃ/m²
= 4/3 √a(4a/ m²)³/₂- m/2 [(4a/m²)²]
= 32a²/3m³ - 8a²/m³
= 8a²/3m³
A perpendicular AC will be drawn to x-axis to make area OABO= area OCABO - area of ΔOCA
applying limit to the equation of =2√ax . dx- mx dx
= 2√a [ x³/₂/ 3/2]⁴ᵃ/m² - m[x²/2]⁴ᵃ/m²
= 4/3 √a(4a/ m²)³/₂- m/2 [(4a/m²)²]
= 32a²/3m³ - 8a²/m³
= 8a²/3m³
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