Length of arc of parabola

$y^2 = 4ax$

diff w.r.t x we get

$2ydy = 4adx$

$\dfrac {dx}{dy}=\dfrac {2y}{4a}$

$\dfrac {dx}{dy}=\dfrac {y}{2a}$

If the line $3y = 8x$ intersects parabola then we have

$x=\dfrac {3y}8,y^2=4ax $ given

$y^2=\dfrac {12ay}8 $ i.e $y=\dfrac {3a}2$

Hence y coordinate of A is $\dfrac {3a}2$

According to formulae

$Arc \space OA = \int\limits_0^{\dfrac {3a}2}\sqrt{1+\Bigg(\dfrac {dx}{dy}\Bigg)^2 dy}$

$= \int\limits_0^{\dfrac {3a}2}\sqrt{1+\dfrac {y^2}{4a^2}dy}$

$= \dfrac 1{2a}\int\limits_0^{\dfrac {3a}2}\sqrt{y^2+4a^2 dy}$

We have direct formulae for this type of integration

Arc OA $=\dfrac 1{2a}\Bigg[\dfrac y2\sqrt{y^24a^2} +\dfrac {4a^2}2\log(y+\sqrt{y^2+4a^2})\Bigg]_0^{\dfrac {3a}2}$

$=\dfrac 1{2a}\Bigg[\dfrac {3a}4\sqrt{\dfrac {9a^2}4} + 4a^2+\dfrac {4a^2}2\log\Bigg(\dfrac {3a}2 +\sqrt{\dfrac {9a^2}4+4a^2}\Bigg)-0-\dfrac {4a^2}2\log(\sqrt{4a^2})\Bigg]$

$=\dfrac 1{2a}\Bigg[\dfrac {3a}4\dfrac {5a}2 +\dfrac {4a^2}2\Bigg[\log \Bigg(\dfrac {3a}2 +\dfrac {5a}2\Bigg) -\log 2a\Bigg]\Bigg]$

$=\dfrac 1{2a}\Bigg[\dfrac {15a^2}8 +2a^2 \Bigg[\log \dfrac {8a}2-\log 2a\Bigg]\Bigg]$

$=\dfrac {a^2}{2a}\Bigg[\dfrac {15}8 +2 \Bigg[\log \dfrac {4a}{2a}\Bigg]\Bigg]$

$=\dfrac {a}{2}\Bigg[\dfrac {15}8 +2 \log 2\Bigg]$

$= a\Bigg[\dfrac {15}{16} + \log 2\Bigg]$