The question is find the arc length of the parabola $y^2 = 4ax$ cut by the line $3y = 8x$ I applied this formula $\int(1+ (dx╱dy)^2) dy $. However by substituting the value of $dx/dy I$ obtain an expression $ 1/2a \int(4a^2 + y^2 ) ^.5 dy$ limits $3a╱2$ to $0$.
I tried integrating this expression by substituting $y = 2a \tan(m)$ but the answer didn't match. What should I do? Where am I going wrong?
I get the following:
The line cutting the parabola is $\;y=\frac83x\;$, so it cuts the parabola whenever
$$\frac{64}9x^2=y^2=4ax\iff 4x\left(\frac{16}9x-a\right)=0\iff x=0\;,\;\;x=\frac{9a}{16}$$
From this it follows at once the line intersects the parabola at the origin and at some point in the first quadrant, and here we have the function
$$y=2\sqrt a\sqrt x\implies y'=\frac{\sqrt a}{\sqrt x}=\sqrt{\frac ax}$$
thus, the wanted length is given by
$$\int\limits_0^{\frac{9a}{16}}\sqrt{1+\frac ax}\;dx\;\ldots$$