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Answer 1

Question:- Find the area common to the circle x² + y² = 16a²  and the parabola y² = 16ax. Hence find the larger of the area into which the circle is divided by the parabola.

Step-by-step explanation:

Given,

Equation of circle,

x² + y² = 16a²     ...................i)

By comparing this equation with equation of circle x² + y² = r²,

We get, r = 4a units.

So area of circle = 16a²π units²

Also equation of parabola,

y² = 6ax   ...............ii)

First of all we need to find the intersection of both conics i.e. circle and parabola.

So putting the value of y² from equation ii) in equation i),

We get,

x² + 6ax  = 16a²

x² + 6ax - 16a² = 0

x² + 8ax - 2ax - 16a² = 0

x(x + 8a) -2a(x + 8a) = 0

(x + 8a)(x - 2a) = 0

∴ x = -8a or x = 2a

By putting  x = -8a in equation ii), we get a imaginary value of y. Therefore this value of x = 8a is negligible.

Now,

Putting  x = 2a in equation ii)

y² = 6a × 2a

y² = 14a²

y = ±a√14

y = ±2a√3

So, intersecting points of given circle and parabola are (2a , 2a√3) and (2a, - 2a√3)

By figure area common to  circle and parabola is OABCO.

By symmetry we can are that area of OABO and OBCO are equal.

Area of OABCO =  2(area of OABO)

Now we will find the area of OABO = S(say

$S=\int\limits^{2a}_0{y_{parabola}\,dx}+\int\limits^{4a}_{2a}{y_{circle}\dx}\\\;$

$S=\int\limits^{2a}_0{\sqrt{6ax}}\,dx+\int\limits^{4a}_{2a}{{\sqrt{16a^2-x^2}\,dx}}$

$S=\sqrt{6a}\int\limits^{2a}_0{\sqrt{x}}\,dx+\int\limits^{4a}_{2a}{{\sqrt{(4a)^2-x^2}\,dx}}\\\;$

$S=\sqrt{6a}[\frac{3}{2}x^{\frac{2}{3}}]\limits^{2a}_{0}+[\frac{x}{2}\sqrt{(4a)^2-x^2}+\frac{1}{2}(4a^2\sin^{-1}\frac{x}{4a})]\limits^{4a}_{2a}\\\;$

$S=\frac{2}{3}\sqrt{6a}[(2a)^\frac{3}{2}-0]+[0+\frac{1}{2}(4a)^2\sin^{-1}\frac{4a}{4a}-(\frac{2a}{2}\sqrt{16a^2-4a^2}+\frac{1}{2}(4a)^2\sin^{-1}\frac{2a}{4a})]\\\;$

$S=\frac{2}{3}(\sqrt{3}.\sqrt{2}.\sqrt{a})[(2\sqrt{2}\times a\sqrt{a})]+[\frac{8a^2\pi}{2}-(\frac{4a^2}{2}\sqrt{3}+\frac{8a^2\pi}{6})]\\\;$

$S=\frac{8\sqrt{3}a^2}{3}+[4a^2\pi-2\sqrt{3}a^2-\frac{4a^2\pi}{3}]\\\;$

$S=\frac{8\sqrt{3}a^2}{3}-2\sqrt{3}a^2+\frac{8a^2\pi}{3}]\\\;$

$S=\frac{2\sqrt{3}a^2}{3}+\frac{8a^2\pi}{3}\\\;$

$S=\frac{2a^2}{3}(\sqrt{3}+4\pi)$

Now,

$\text{Area of OABCO}=2\times S=\frac{4a^2}{3}(\sqrt{3}+4\pi)$

Area of larger section divided by Parabola,

$=16a^2\pi-\frac{4a^2}{3}(\sqrt{3}+4\pi)\\\;\\=16a^2\pi-\frac{4\sqrt{3}a^2}{3}-\frac{16a^2\pi}{3}\\\;\\=\frac{32a^2\pi}{3}-\frac{4\sqrt{3}a^2}{3}\\\;\\=\frac{4a^2}{3}(8\pi-\sqrt{3})$

Answer 2

Step-by-step explanation:

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