Question

I the area bounded by the parabola y^2=16ax and the line y=4mx is a^2/12 sq.Units then using integration find the value of m

Answer 1

first of all, find intersecting points of curves y = 4mx and y² = 16ax

=> (4mx)² = 16ax

=> 16m²x² = 16ax

=> m²x² - ax = 0

=> x(m²x - a) = 0

=> x = 0, a/m²

now, area bounded by the curves = $\int\limits^{a/m^2}_0{\sqrt{16ax}-4mx}\,dx$

= $\sqrt{16a}\int\limits^{a/m^2}_0{\sqrt{x}}\,dx-\int\limits^{a/m^2}_0{4mx}\,dx$

= $\sqrt{16a}\left[\frac{2x^{3/2}}{3}\right]^{a/m^2}_0-[2mx^2]^{a/m^2}_0$

= √(16a) × 2(a/m²)^(3/2)/3 - 2m × a²/m⁴

= 4√a × 2a√a/3m³ - 2a²/m³

= 8a²/3m³ - 2a²/m³

= (8a² - 6a²)/3m³

= 2a²/3m³

but given area enclosed by curves = a²/12

so, a²/12 = 2a²/3m³

or, 1/12 = 2/3m³

or, 1/4 = 2/m³

or, m³ = 8

hence, m = 2

Answer 2

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Step-by-step explanation: attachment