Question

If $A_1$ is the area of the parabola $y^2=4ax$ lying between vertex and the latus rectum and $A_2$ is the area between the latus rectum and the double ordinate $x=2a$, then $\frac{A_1}{A_2}=$

• A. $2\sqrt{2}-1$
• B. $\frac{1}{7}(2\sqrt{2}+1)$
• C. $\frac{1}{7}(2\sqrt{2}-1)$
• D. None of these

Answer 1

Answer: (B)

$\frac{1}{7}(2\sqrt{2}+1)$

The areas $A_1$ and $A_2$ are shown below:

Now, $A_1=2{\int }_0^a\sqrt{4ax}dx=\frac{8a^2}{3}$ and $A_2=2\left[{\int }_0^{2a}\sqrt{4ax}-{\int }_0^a\sqrt{4ax}\right]=\frac{16\sqrt{2}}{2}{a}^2-\frac{8a^2}{3}$

$=\frac{8a^2}{3}(2\sqrt{2}-1)\Rightarrow \frac{A_1}{A_2}=\frac{1}{2\sqrt{2}-1}=\frac{2\sqrt{2}+1}{7}$