Question

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

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

Answer 1

The correct option is A

$6(\sqrt{2}+\log \left(\sqrt{2}+1\right))$

Find the length of the parabola

The equation of the parabola is $y^2=12x$ $...\left(i\right)$

Therefore the equation for the latus rectum is $x=3$ $...\left(ii\right)$

From equations $\left(i\right),\left(ii\right)$ we get

$y^2=36$

$\Rightarrow$ $y=\pm 6$

Hence the end points of the latus rectum are $\left(3,6\right)$ and $(3,-6)$

Differentiate $\left(i\right)$ with respect to $x$

$2y\frac{dy}{dx}=12$

$\Rightarrow$ $\frac{dy}{dx}=\frac{6}{y}$

The required length is given by

Length $=2{\int }_0^3\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx$

$=2{\int }_0^3\sqrt{1+{\left(\frac{6}{y}\right)}^2}dx$

$=2{\int }_0^3\sqrt{\frac{y^2+36}{y^2}}dx$

$=2{\int }_0^3\sqrt{\frac{12x+36}{12x}}dx$

$=2{\int }_0^3\frac{x+3}{\sqrt{x^2+3x}}dx$

$=2{{\left[\sqrt{x^2+3x}+\frac{3}{2}\log \left|\left(x+\frac{3}{2}\right)+\sqrt{x^2+3x}\right|\right]}_0}^3$

$=2\left[3\sqrt{2}+\frac{3}{2}\log \left(\frac{9}{2}+3\sqrt{2}\right)-\frac{3}{2}\log \left(\frac{3}{2}\right)\right]$

$=2\left(3\sqrt{2}+3\log (\sqrt{2}+1)\right)$

$=6(\sqrt{2}+\log \left(\sqrt{2}+1\right))$

So, the length of the parabola cut off by the latus rectum is $6(\sqrt{2}+\log \left(\sqrt{2}+1\right))$

Hence option $\left(A\right)$ is the correct answer.