The normal to the curve y²=c e^(x/a)+4a(x+a) at a point M

The normal to the curve $y^{2}=c\,e^{x/a}+4a(x+a)$ at a point $M$ meets the $Ox$ axis at the point $S$. Prove that the midpoint of the segment $\overline{MS}$ lies on the parabola $y^{2}=ax$.

Step 1: Differentiate implicitly

Let $M=(x_{0},y_{0})$ be on the curve, so $y_{0}^{2}=c\,e^{x_0/a}+4a(x_{0}+a)$. Differentiating, \[2y\,y' = \frac{c}{a}e^{x/a}+4a \;\Longrightarrow\; y'(x_{0}) = \frac{\dfrac{c}{a}e^{x_0/a}+4a}{2y_{0}}. \tag{1}\]

Step 2: Normal at $M$, and where it meets $Ox$

The normal is $y-y_{0}=-\dfrac{1}{y'(x_0)}(x-x_{0})$; setting $y=0$ (assuming $y_{0}\neq0$) and solving for $x_{S}$, \[x_{S}-x_{0} = \frac{\dfrac{c}{a}e^{x_0/a}+4a}{2} \;\Longrightarrow\; x_{S} = x_{0}+\frac{\dfrac{c}{a}e^{x_0/a}+4a}{2}. \tag{2}\]

Step 3: Midpoint $T$ of $\overline{MS}$

\[ \eta = \frac{y_{0}}{2}, \qquad \xi = \frac{x_{0}+x_{S}}{2} = x_{0} + \frac{\dfrac{c}{a}e^{x_0/a}+4a}{4}. \tag{3} \]

Step 4: Eliminate the exponential using the curve equation

From $y_{0}^{2}=c\,e^{x_0/a}+4a(x_0+a)$, $\dfrac{c}{a}e^{x_0/a}=\dfrac{y_{0}^{2}-4a(x_0+a)}{a}$; substituting into (3), the $a^2$ terms cancel and

\[ \xi = x_{0} + \frac{y_{0}^{2}-4ax_{0}}{4a} = \frac{4ax_{0}+y_{0}^{2}-4ax_{0}}{4a} = \frac{y_{0}^{2}}{4a}. \tag{4} \]

Step 5: $T$ satisfies $y^{2}=ax$

Since $y_{0}=2\eta$, (4) gives $\xi=\dfrac{4\eta^{2}}{4a}=\dfrac{\eta^{2}}{a}$, i.e.

\[ \eta^{2} = a\xi. \tag{5} \]

Step 6: Verification

\[(a,c,x_0)=(1,2,0.5)\text{: } \eta^2-a\xi=0 \text{ (numerically)}\ \checkmark\]

\[(a,c,x_0)=(2,1,1.0),\ (0.5,3,0.2)\text{: } \eta^2-a\xi=0\ \checkmark\]