Question

If the area bounded by the parabola $y^2=4ax$ and the double ordinate $x=6$ is twice the area bounded by the parabola and its latus rectum, then the length of the latus rectum is

• A. $6\sqrt{2}$
• B. $12\sqrt{2}$
• C. $6\left(3^{\frac{1}{3}}\right)$
• D. $12\left(2^{\frac{1}{3}}\right)$

Answer 1

The correct option is D $12\left(2^{\frac{1}{3}}\right)$

From the above question,
Area bounded by the parabola and its latus rectum
$=2{\int }_0^{6}y.dx$
$=2\times 2{\int }_0^{a}y.dx$
$\Rightarrow {\int }_0^6\sqrt{x}dx=2{\int }_0^a\sqrt{x}dx$
$\Rightarrow {\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^6=2{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^a$
$\Rightarrow \frac{2}{3}\times 6\sqrt{6}=2\times \frac{2}{3}a\sqrt{a}$
On simplification,we get
$a^{\frac{3}{2}}=3\sqrt{6}$
$\Rightarrow a={\left(3\sqrt{6}\right)}^{\frac{2}{3}}$
$=3^{\frac{2}{3}}{6}^{\frac{1}{3}}$
Using ${\left(ab\right)}^m=a^m{b}^m$
$=3^{\frac{2}{3}}{3}^{\frac{1}{3}}{2}^{\frac{1}{3}}$
Using $a^m{a}^n=a^{m+n}$ we get
$=3\left(2^{\frac{1}{3}}\right)$
$\therefore$ Length of latus rectum$=4a=4\times 3\left(2^{\frac{1}{3}}\right)=12\left(2^{\frac{1}{3}}\right)$