Question
If the area enclosed by $${y^2} = 4ax$$ is $$\frac{1}{3}\,sq.$$ unit, then the roots of the equation $${x^2} + 2x = a,$$ are :
A.
$$ - 4{\text{ and }}2$$
B.
$$4{\text{ and }}2$$
C.
$$ - 2{\text{ and }} - 4$$
D.
$$8{\text{ and }} - 8$$
Answer :
$$ - 4{\text{ and }}2$$
Solution :
$$\eqalign{
& y = \int_0^{\frac{4}{a}} {\left( {a.x - \sqrt {4a.x} } \right)dx} \cr
& \frac{1}{3} = \int_0^{\frac{4}{a}} {ax\,dx - \int_0^{\frac{4}{a}} {\sqrt {4ax} \,dx} } \cr
& \frac{1}{3} = \left[ {\frac{{a{x^2}}}{2}} \right]_0^{\frac{4}{a}} - 2\left[ {\frac{{{{\left( {4ax} \right)}^{\frac{3}{2}}}}}{3}} \right]_0^{\frac{4}{a}} \cr
& \frac{1}{3} = \frac{{\frac{{16a}}{{{a^2}}}}}{2} - \frac{2}{3}\left[ {4a{{\left( {\frac{4}{a}} \right)}^{\frac{3}{2}}}} \right],\,a = 8 \cr} $$
Putting the value of $$a$$ in $${x^2} + 2x - a = 0,$$ we get its roots i.e., $$ – 4$$ and $$2.$$