Question

Assertion : The area bounded by the curves $y^2=4a^2(x-1)$ and lines $x=1$ and $y=4a$ is $\frac{16a}{3}$ sq. units.
Reason : The area enclosed between the parabola $y^2=x^2-x+2$ and the line $y=x+2$ is $\frac{8}{3}$ sq. units.

• A. Assertion is correct, Reason is correct; Reason is a correct explanation for assertion.
• B. Assertion is correct, Reason is correct; Reason is not a correct explanation for Assertion
• C. Assertion is correct, Reason is incorrect
• D. Assertion is incorrect, Reason is correct.

Answer 1

Answer: (C)

Assertion : On solving $y^2=4a^2(x-1)$ and $y=4a$, we get $x=5$

Required area $=\underset{1}{\overset{5}{\int }}(4a-2a\sqrt{x-1})dx$
$={\left[4ax-2a\cdot \frac{(x-1{)}^{3/2}}{3/2}\right]}_1^5=\frac{16a}{3}sq$.units
Reason : Given equation of parabola can be rewritten as

${\left(x-\frac{1}{2}\right)}^2=y-\frac{7}{4}$
Parabola $y^2=x^2-x+2$ and the line $y=x+2$ intersects at the point $(0,0)$ and $(2,4)$
$\therefore$ Required area $=\underset{0}{\overset{2}{\int }}\left[(x+2)-\left(x^2-x+2\right)\right]dx$
$=\underset{0}{\overset{2}{\int }}\left(-x^2+2x\right)dx={\left[\frac{-x^3}{3}+x^2\right]}_0^2=\frac{4}{3}$ sq.units