Question

Find the area bounded by the curves y² = 4a² (x -1) and the line x =1, y = 4a.

Answer 1

Answer:

$\frac{28a}{3}$ Sq. Units.

Step-by-step explanation:

We have the equation of curve as y²=4a²(x-1) ........ (1)

⇒ $(x-1)=\frac{y^{2} }{4a^{2} }$

⇒ $x=(\frac{y^{2} }{4a^{2} }+1 )$ .......... (2)

Now, we have to find the area bounded by curve (1) and the lines x=1 and y=4a.

From equation (1), at x=1, y=4a²(0)=0 can be written.

So, by integrating the curve (2), from y=0 to y=4a limits, we can get the required area.

First, we will calculate the indefinite integral then we will put the limits to get the area.

Hence,

∫$(\frac{y^{2} }{4a^{2} }+1 )dy$

= $\frac{1}{4a^{2} }(\frac{y^{3} }{3} ) +y+c$ {Where c is an integration constant}

Now, putting the limits we get,

Area= $\frac{1}{4a^{2} } [\frac{y^{3} }{3} ]_{0} ^{4a}+  [y ]_{0} ^{4a}$

       =$\frac{1}{4a^{2} }.\frac{64a^{3} }{3} +4a$

       =$\frac{16a}{3}+4a$

       =$\frac{28a}{3}$ Sq. Units. (Answer)

Answer 2

Answer

y^2=4a^2(x-1).......(i)

Line x=1

Line y=4a

Put y=4a in (i)

At y=4a,we get x=5.

Now,

Area=integration of (4a-2a×squareroot of x-1)from x=0 to x=5

On solving we get:

16a/3 sq units