Prove that the parabolas y²=4ax+4a² and y²=-4bx+4b²

Prove that the parabolas $y^{2}=4ax+4a^{2}$ and $y^{2}=-4bx+4b^{2}$ ($a,b>0$) intersect at right angles.

Step 1: Intersection points

Equating the two expressions for $y^2$: $4ax+4a^2=-4bx+4b^2 \Rightarrow x=b-a$; substituting back, \[y^{2}=4a(b-a)+4a^{2}=4ab \;\Longrightarrow\; y=\pm2\sqrt{ab}. \tag{1}\]

Step 2: Slopes at an intersection point

Implicit differentiation gives $y'=2a/y$ for the first parabola and $y'=-2b/y$ for the second; at $y=\pm2\sqrt{ab}$, \[k_{1}=\frac{2a}{\pm2\sqrt{ab}}=\pm\sqrt{\frac ab}, \qquad k_{2}=\frac{-2b}{\pm2\sqrt{ab}}=\mp\sqrt{\frac ba}. \tag{2}\]

Step 3: Perpendicularity

\[ k_{1}k_{2} = \pm\sqrt{\frac ab}\cdot\!\left(\mp\sqrt{\frac ba}\right) = -\sqrt{\frac ab\cdot\frac ba} = -1. \tag{3} \]