Question

The area bounded by the curve $y^2 = 4a^2 ( x - 1)$ and the lines $x = 1$ and $y = 4a$ is equal to :

• A. $\frac{16a}{3}$ sq unit
• B. $5a$ sq unit
• C. $\frac{17a}{4}$ sq unit
• D. None of these

Answer 1

Answer: (A)

Given curve : $y^2 = 4a^2 (x - 1)$
Lines : $x = 1$ and $y = 4a$
Now, $(y - 0)^2 = 4a^2 (x - 1)$
This is a parabola with vertex $A (1, \,0)$.

Required area
= area of shaded region ABC
$= \int\limits^{5}_{1}$ [y(line) - y(parabola)]dx
$= \int\limits^{5}_{1} \left[4a-2a\sqrt{x-1}\right]dx$
$= \int\limits^{5}_{1}\left[4ax-2a\times\frac{2}{3}\left(x-1\right)^{3 2}\right]^{5}_{1}$
$= \left(20a-\frac{32a}{3}\right) - \left(4a-0\right) = \frac{16}{3}a$ sq.unit