On the curve xy²=2a³ find the point whose normal passes

On the curve $xy^{2}=2a^{3}$ find the point whose normal passes through the origin.

Step 1: Implicit differentiation

Differentiating $xy^{2}=2a^{3}$ with respect to $x$: $y^{2}+2xy\,y'=0 \Rightarrow y'=-\dfrac{y}{2x}$. The slope of the normal at $(x_{0},y_{0})$ is

\[ m = -\frac{1}{y'} = \frac{2x_{0}}{y_{0}}. \tag{1} \]

Step 2: Condition that the normal pass through the origin

The normal at $P=(x_{0},y_{0})$ is $y-y_{0}=\dfrac{2x_{0}}{y_{0}}(x-x_{0})$; substituting $(0,0)$: \[-y_{0} = -\frac{2x_{0}^{2}}{y_{0}} \;\Longrightarrow\; y_{0}^{2} = 2x_{0}^{2}. \tag{2}\]

Step 3: Solve with the curve equation

Combining (2) with $x_{0}y_{0}^{2}=2a^{3}$: \[x_{0}\cdot2x_{0}^{2}=2a^{3} \;\Longrightarrow\; x_{0}^{3}=a^{3} \;\Longrightarrow\; x_{0}=a, \qquad y_{0}^{2}=2a^{2} \;\Longrightarrow\; y_{0}=\pm a\sqrt2. \tag{3}\]

(Real $a$: $x_0^3=a^3$ has the single real root $x_0=a$, the other two cube roots being complex.)

Step 4: Verification

\[\text{On the curve: } a\cdot(a\sqrt2)^2=2a^3\ \checkmark\]

\[\text{Normal at } P_1\text{: slope } m=\tfrac{2a}{a\sqrt2}=\sqrt2;\ y-a\sqrt2=\sqrt2(x-a)\Rightarrow y=\sqrt2\,x;\ \text{passes through } O\ \checkmark\]

\[\text{By symmetry } y\to-y\text{: the normal at } P_2 \text{ is } y=-\sqrt2\,x,\text{ also through } O\ \checkmark\]