Question

Find the area of the region bounded by the curve y^2 = 4ax and the line y= mx

Answer 1

Answer: 8a²/3m³

Step-by-step explanation:

Given,

Equation of Parabola,

y² = 4ax  ..............i)

Equation of Line,

y = mx  ............ii)

First we need to find intersecting points of Parabola and Line:-

Putting y = mx in equation i)

(mx)² = 4ax

m²x² = 4ax

m²x² - 4ax = 0

x(m²x - 4a) = 0

x = 0 and  x = 4a/m²

When x = 0, y = 0

When x = 4a/m²  , y = 4a/m

Hence,

Intersecting points of Parabola and line are A(0,0) and B(4a/m²,4a/m)

Now,

Area ADBCA(Closed by Parabola and Line);

$=\int\limits^\frac{4a}{m^2}_0{y_{curve}}\,dx\\\;\\=\int\limits^\frac{4a}{m^2}_0{\sqrt{4ax}}\,dx\\\;\\=\sqrt{4a}\int\limits^\frac{4a}{m^2}_0{\sqrt{x}}\,dx\\\;\\=2\sqrt{a}[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}]\limits^\frac{4a}{m^2}_0\\\;\\=\frac{4}{3}\sqrt{a}[(\frac{4a}{m^2})^{\frac{3}{2}}-0]\\\;\\=\frac{4}{3}\sqrt{a}[\frac{4a}{m^2}\sqrt{\frac{4a}{m^2}}]\\\;\\=\frac{16a\sqrt{a}}{3m^2}[\frac{2\sqrt{a}}{m}]\\\;\\=\frac{32a^2}{3m^3}$

Similarly,

Area AEBCA (Closed by line and x-axis);

$=\int\limits^\frac{4a}{m^2}_0{y_{Line}}\,dx\\\;\\=\int\limits^\frac{4a}{m^2}_0{mx}\,dx\\\;\\=m\int\limits^\frac{4a}{m^2}_0{x}\,dx\\\;\\=m[\frac{x^2}{2}]\limits^\frac{4a}{m^2}_0\\\;\\=\frac{m}{2}[(\frac{4a}{m^2})^2-0]\\\;\\=\frac{m}{2}[\frac{16a^2}{m^4}]\\\;\\=\frac{8a^2}{m^3}$

Now,

We have to find area ADBEA

Area of ADBEA = area of ADBCA - area of ADBEA

(Refer to attachment)

Area of ADBEA =

$\frac{32a^2}{3m^3}-\frac{8a^2}{m^3}\\\;\\=\frac{a^2}{m^3}(\frac{32}{3}-\frac{8}{1})\\\;\\=\frac{a^2}{m^3}(\frac{32-24}{3})\\\;\\=\frac{8a^2}{3m^3}$

Which is the required area