A normal tangent

Solving for $y$ , I get $y=\pm x^{3/2}$ . This relation is only defined in the real numbers for $x\geq 0$ . Since $x^3=4m^2$ is bijective for $m,x>0$ , we can replace $x$ with $4m^2$ . So, then the relation can be expressed as $y=\pm (4m^2)^{3/2}=\pm 8m^3$ , giving us the parametrization $x=4m^2,y=\pm 8m^3, m\geq 0$ . Next, find the derivative: $\frac{dy}{dx}=\frac{dy/dm}{dx/dm}=\frac{\pm 24m^2}{8m}=\pm 3m$ . We are looking for two points on the curve such that the tangent line of one is the normal line to the other. This means, first of all, that the product of their slopes must be -1. Since $m\geq 0$ , we need to take one point from the positive half of the curve, and another from the negative half in order to get a negative result (when multiplying their tangent slopes). So, we need $(3m_1)(-3m_2)=-1$ or $m_1m_2=\frac{1}{9}$ . We also need $8m_2^3=3m_1(4m_2^2-4m_1^2)+8m_1^3$ (by the point-slope formula). I will make a substitution from the first equation into the second into order to find $m_1$ . ( $m_1$ is the one we want to know because $m_1$ lies on $(4m^2,8m^3)$ , which is the one the question is asking for. Whereas, $m_2$ lies on $(4m^2,-8m^3)$ , which is not the desired part of the curve.) So, now to solve: $8m_2^3=3m_1(4m_2^2-4m_1^2)+8m_1^3 \Rightarrow 2m_2^3-2m_1^3=3m_1(m_2^2-m_1^2)$ $\Rightarrow 2(m_2-m_1)(m_1^2+m_1m_2+m_2^2)=3m_1(m_2-m_1)(m_2+m_1)$ $\Rightarrow 2(m_1^2+\frac{1}{9}+m_2^2)=3m_1(m_2+m_1) \Rightarrow 2(m_1^2+\frac{1}{9}+\frac{1}{(9m_1)^2})=3(\frac{1}{9}+m_1^2)$ $\Rightarrow 81m_1^4+9m_2^2-2=0 \Rightarrow m_1=\frac{\sqrt{2}}{3}$ .