Question

The length of the arc of the parabola $x^2=4ay$ measured from the vertex to one extremity of the Latus-Rectum is:

• A. $a[\sqrt{2}+\log (1-\sqrt{2})]$
• B. $a[\sqrt{2}+\log (1+\sqrt{2})]$
• C. $a[\sqrt{2}-\log (1+\sqrt{2})]$
• D. $a[\sqrt{2}-\log (1-\sqrt{2})]$

Answer 1

The correct option is B $a[\sqrt{2}+\log (1+\sqrt{2})]$

Let $A$ be the vertex and $L$ an extremity of the Latus-Rectum so that at $A,x=0$ and at $L,x=2a.$
Now, $y=\frac{x^2}{4a}$ so that $\frac{dy}{dx}=\frac{1}{4a}.2x=\frac{x}{2a}$
$\therefore$arc$AL={\int }_0^{2a}\sqrt{(1+{\left(\frac{dy}{dx}\right)}^2)}dx$
$={\int }_0^{2a}\sqrt{(1+{\left(\frac{x}{2a}\right)}^2)}dx$
$=\frac{1}{2a}{\int }_0^{2a}\sqrt{{\left(2a\right)}^2+x^2}dx$
$=\frac{1}{2a}{[\frac{x\sqrt{{\left(2a\right)}^2+x^2}}{2}+\frac{{\left(2a\right)}^2}{2}\sin h^{-1}\left(\frac{x}{2a}\right)]}_0^{2a}$
$=\frac{1}{2a}[\frac{2a\sqrt{{\left(8a\right)}^2}}{2}+2a^2\sin h^{-1}1]$
Since $\sin h^{-1}x=\log (x+\sqrt{1+x^2})$ we have
Length of arc AL$=a[\sqrt{2}+\log (1+\sqrt{2})]$