Question

The parabolas $y^2=4x,x^2=4y$ divide the square region bounded by the lines $x=4,y=4$ and the coordinates axes. If $A_1,A_2,A_3$ are respectively the areas of these parts numbered top to bottom then $\frac{A_1}{A_2}+\frac{A_2}{A_3}+\frac{A_3}{A_1}=$.......

• A. $0$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

$3$

Area A_2 is common area between parabolas

$\Rightarrow A_2={\int }_0^4(2\sqrt{x}-\frac{x^2}{4})dx$

$\Rightarrow 2[\frac{x^{3/2}}{3/2}{]}_0^4-[\frac{x^3}{12}{]}_0^4$

$\Rightarrow 2\times \frac{4^{3/2}}{3/2}-\frac{4^3}{12}$

$\Rightarrow \frac{32}{3}-\frac{16}{3}=\frac{16}{3}$

Therefore, $A_2=\frac{16}{3}$

Now, $A_1+A_3=(4\times 4)-\frac{16}{3}$

$\Rightarrow A_1+A_3=16-\frac{16}{3}=\frac{32}{3}$

$\Rightarrow A_1=A_3=\frac{1}{2}\times \frac{32}{3}=\frac{16}{3}$

Therefore, $A_1=A_2=A_3=\frac{16}{3}$

Hence, $\frac{A_1}{A_2}+\frac{A_2}{A_3}+\frac{A_3}{A_1}=1+1+1=3$