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

Answer: (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,
$ds=\sqrt{{\left(dx\right)}^2+{\left(dy\right)}^2}=\sqrt{{\left(\frac{dx}{dy}\right)}^2+1dy}$
$\Rightarrow s={\int }_{-6}^6\left[\sqrt{{\left(\frac{dx}{dy}\right)}^2+1}\right].dy...\left(1\right)$
From $y^2=12x\Rightarrow x=\frac{y^2}{12}<br/>\frac{dx}{dy}=\frac{2y}{12}=\frac{y}{6}$ Putting in $\left(1\right)$,
$s={\int }_{-6}^6\left[\sqrt{{\left(\frac{y}{6}\right)}^2}+1\right]dy=2{\int }_0^6\sqrt{\frac{y^2+36}{36}dy}$
$=\frac{2}{6}{\int }_0^6\sqrt{y^2+6^2}dy$
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$=\frac{1}{3}{\left[\frac{y}{2}\sqrt{6^2+y^2}+\frac{6^2}{2}log\left(y+\sqrt{6^2+y^2}\right)+C\right]}_{{}_{{}_0}}^{{}^6}$
$=\frac{1}{3}\left[\frac{6}{2}\sqrt{6^2+6^2}+18log\left(6+\sqrt{6^2+6^2}\right)+C-0-18log\left(0+\sqrt{6^2+0}\right)-C\right]$
$=\frac{1}{3}\left[3.6\sqrt{2}+18log\left(6+6\sqrt{2}\right)-18log6\right]$
$=6\sqrt{2}+6log\frac{6\left(1+\sqrt{2}\right)}{6}$
$=6\left[\left(\sqrt{2}+log\left(1+\sqrt{2}\right)\right)\right]$