Tangents to the curves ay = x² and y = x²+x, drawn

Tangents to the curves $ay = x^{2}$ and $y = x^{2}+x$, drawn at points with the same abscissa, intersect at a point $S$. Determine $a$ so that $S$ lies on the line $y = 2x$.

Step 1: The two tangents at abscissa $x_{0}$

For $y = \dfrac{x^{2}}{a}$: slope $\dfrac{2x_{0}}{a}$, so

\[ y = \frac{2x_{0}}{a}x-\frac{x_{0}^{2}}{a}. \tag{1} \]

For $y = x^{2}+x$: slope $2x_{0}+1$, so

\[ y = \bigl(2x_{0}+1\bigr)x-x_{0}^{2}. \tag{2} \]

Step 2: Their intersection

Equating (1) and (2) and multiplying by $a$: \[\Bigl(2x_{0}\bigl(1-a\bigr)-a\Bigr)x = x_{0}^{2}\bigl(1-a\bigr) \;\Longrightarrow\; x_{S} = \frac{x_{0}^{2}\bigl(1-a\bigr)}{2x_{0}\bigl(1-a\bigr)-a}. \tag{3}\]

Step 3: Impose $y_{S} = 2x_{S}$

From (2), $y_{S} = \bigl(2x_{0}+1\bigr)x_{S}-x_{0}^{2}$, so the condition reads

\[ \bigl(2x_{0}+1\bigr)x_{S}-x_{0}^{2} = 2x_{S} \;\Longrightarrow\; \bigl(2x_{0}-1\bigr)x_{S} = x_{0}^{2}. \tag{4} \]

Substituting (3) and cancelling $x_{0}^{2}\neq0$: $\bigl(2x_{0}-1\bigr)\bigl(1-a\bigr) = 2x_{0}\bigl(1-a\bigr)-a$.

The terms $2x_{0}\bigl(1-a\bigr)$ cancel, leaving

\[ -\bigl(1-a\bigr) = -a \;\Longrightarrow\; 2a = 1 \;\Longrightarrow\; a = \frac12. \tag{5} \]

No value of $x_{0}$ survives: the condition holds for every common abscissa at once.

Step 4: Verification

\[a = \tfrac12:\ \text{the curves are } y = 2x^{2} \text{ and } y = x^{2}+x,\ \text{with tangents } y = 4x_{0}x-2x_{0}^{2} \text{ and } y = \bigl(2x_{0}+1\bigr)x-x_{0}^{2}\]

\[\bigl(2x_{0}-1\bigr)x = x_{0}^{2} \;\Longrightarrow\; x_{S} = \frac{x_{0}^{2}}{2x_{0}-1}, \quad y_{S} = \frac{2x_{0}^{2}}{2x_{0}-1} = 2x_{S} \;\;\checkmark\]

\[x_{0} = 1:\ S = \bigl(1,2\bigr); \qquad x_{0} = 2:\ S = \left(\tfrac43,\tfrac83\right) \;\;\checkmark\ \text{(both on } y = 2x)\]