Question

The length of the parabola y$^2$ = 12x cut off by the latus-rectum is

• A. $6\left(\sqrt{2}+log\left(1+\sqrt{2}\right)\right)$
• B. $3\left(\sqrt{2}+log\left(1+\sqrt{2}\right)\right)$
• C. $6\left(\sqrt{2}-log\left(1+\sqrt{2}\right)\right)$
• D. $3\left(\sqrt{2}-log\left(1+\sqrt{2}\right)\right)$

Answer 1

The Correct Option is A

On comparing the equation of the parabola
y$^2$ = 12 x with the standard equation,
y$^2$ = 4 ax, we get 4 a = 12 or a = 3.
Hence, the focus, point C will be at (3, 0) and the extremities of the latus-rectum AB will be at (a, 2a) and (a, -2a). So the coordinates of A and B are (3, 6) and (3, -6) respectively. Now we need to find the length ( curve AOB) of the parabola. As it is not a straight line so we cannot directly find the length of this curve as we cannot directly apply Pythagorous theorem. Let us consider a small length ds on the parabola. Using pythagorous theorem for this length,
$ds=\sqrt{\left(dx\right)^{2} +\left(dy\right)^{2}}=\sqrt{\left(\frac{dx}{dy}\right)^{2} +1 dy}$
$\Rightarrow s = \int^{6}_{ -6}\left[\sqrt{\left(\frac{dx}{dy}\right)^{2} +1}\right].dy \quad\quad\quad...\left(1\right)$
From $y^{2} = 12 x \Rightarrow x=\frac{y^{2}}{12}
\frac{dx}{dy}=\frac{2y}{12}=\frac{y}{6}\quad$ Putting in $\left(1\right)$,
$s=\int^{6}_{-6}\left[\sqrt{\left(\frac{y}{6}\right)^{2}}+1\right]dy =2\int^{6}_{0}\sqrt{\frac{y^{2} +36}{36}dy}$
$=\frac{2}{6}\int^{6}_{0}\sqrt{y^{2}+6^{2}}dy$
Using $\int \sqrt{x^{2}+a^{2}}=\frac{x}{2}\sqrt{a^{2}+x^{2}}$
$+\frac{a^{2}}{2}log\left(x+\sqrt{a^{2}+x^{2}}\right)+C$
We get s$=\frac{1}{3}\left[\frac{y}{2}\sqrt{6^{2}+y^{2}}+\frac{6^{2}}{2}log\left(y+\sqrt{6^{2}+y^{2}}\right)+C\right]^{^6}_{_{_0}}$
$=\frac{1}{3}\left[\frac{6}{2}\sqrt{6^{2}+6^{2}}+18 log \left(6+\sqrt{6^{2}+6^{2}}\right)+C-0-18 log \left(0+\sqrt{6^{2}+0}\right)-C\right]$
$=\frac{1}{3}\left[3.6\sqrt{2}+18log\left(6+6\sqrt{2}\right)-18log 6\right]$
$=6\sqrt{2}+6log\frac{6\left(1+\sqrt{2}\right)}{6}$
$=6\left[\left(\sqrt{2}+log\left(1+\sqrt{2}\right)\right)\right]$