Question

Find the length of the arc of the cal x square is equal to eat from the vertex to an activity of a letter rectum

Answer 1

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

Latus-Rectum is a chord that passes through a focus and is parallel to the directrix

Let A be the vertex and L an extremity of the Latus-Rectum so that at A,x=0 and at L,x=2a.

Then:

$y=x^2/(4a),y'=x/(2a)$

focus: $(-a,0)$

directrix: $x=a$

arc $AL=\int^{2a}_0\sqrt{1+(y')^2}dx=\int^{2a}_0\sqrt{1+(\frac{x}{2a})^2}dx=$

$x=2asinhu\to dx=2acoshudu$

$=\int 2acoshu\sqrt{4a^2sinh^2u+4a^2}du=4a^2\int cosh^2udu=$

$=\frac{coshusinhu+u}{2}=(\frac{x\sqrt{(x/(2a))^2+1}}{2}+sinh^{-1}\frac{x}{2a})|^{2a}_0=a(\sqrt 2+sinh^{-1}1)$