Question

Let y=f(x) be the given curve and $x=a$, $x=b$ be two ordinates then area bounded by the curve $y=f\left(x\right)$, the axis of x between the ordinates $x=a$ & $x=b$, is given by definite integral
${\int }_a^{b}ydx$ or ${\int }_a^{b}f\left(x\right)dx$ and the area bounded by the curve $x=f\left(y\right)$, the axis of y & two abscissae $y=c$ & $y=d$ is given by ${\int }_c^{d}xdy$ or ${\int }_c^{d}f\left(x\right)dy$. Again if we consider two curves $y=f\left(x\right)$, $y=g\left(x\right)$ where $f\left(x\right)\ge g\left(x\right)$ in the interval [a, b] where $x=a$ & $x=b$ are the points of intersection of these two curves Shown by the graph given
Then area bounded by these two curves is given by
${\int }_a^b[f\left(x\right)-g\left(x\right)]dx$
On the basis of above information answer the following questions.

If the area enclosed by the parabola $y^2=64x$ & its latus rectum is $\lambda$, then value of $3\lambda$ equals

• A. $2048$
• B. $128$
• C. $64$
• D. $32$

Answer 1

The correct option is A $2048$

Given curve is $y^2=64x=4\left(16\right)x=4ax$ whose latus rectum is $x=a=16$
$\therefore$ Required area is given by $\lambda$
$\lambda =2{\int }_0^{16}ydy=2{\int }_0^{16}8\sqrt{x}dx$
$=2\times 8\times \frac{2}{3}{\left(x^{3/2}\right)}_0^{16}$, $\lambda =\frac{32\left(64\right)}{3}$
$\therefore$ $3\lambda =32\times 64=2048$