Find the parabola with axis parallel to Oy that passes

Find the parabola with axis parallel to $Oy$ that passes through $\bigl(1,2\bigr)$ and $\bigl(-1,6\bigr)$ and is tangent to the line $2x-y-1 = 0$.

Step 1: The unknowns

\[ y = ax^{2}+bx+c, \qquad a\neq0. \]

Step 2: The two point conditions

\[ a+b+c = 2, \qquad a-b+c = 6. \]

Subtracting gives $2b = -4$, and adding gives $2a+2c = 8$: \[b = -2, \qquad c = 4-a. \tag{1}\]

Step 3: The tangency condition

The line is $y = 2x-1$, of slope $2$. At the point of contact $x_{1}$, the slopes agree: \[2ax_{1}+b = 2 \;\Longrightarrow\; 2ax_{1} = 4 \;\Longrightarrow\; x_{1} = \frac{2}{a}. \tag{2}\] and the point lies on both curves. On the line, $y_{1} = 2x_{1}-1 = \dfrac4a-1$; on the parabola, using (1) and (2), $a\cdot\frac{4}{a^{2}}-2\cdot\frac{2}{a}+\bigl(4-a\bigr) = \frac4a-\frac4a+4-a = 4-a$.

Equating the two values of $y_{1}$: \[\frac4a-1 = 4-a \;\Longrightarrow\; 4-a = 4a-a^{2} \;\Longrightarrow\; a^{2}-5a+4 = 0 \;\Longrightarrow\; \bigl(a-1\bigr)\bigl(a-4\bigr) = 0. \tag{3}\]

Step 4: The two parabolas

\[ a = 1,\ c = 3:\quad y = x^{2}-2x+3, \qquad\text{touching at } \bigl(2,3\bigr); \]

\[ a = 4,\ c = 0:\quad y = 4x^{2}-2x, \qquad\text{touching at } \left(\frac12,\,0\right). \]

Step 5: Verification

\[y = x^{2}-2x+3:\ y(1) = 2,\ y(-1) = 6,\ y'(2) = 2,\ \text{and } 2\cdot2-3-1 = 0 \;\;\checkmark\]

\[y = 4x^{2}-2x:\ y(1) = 2,\ y(-1) = 6,\ y'\!\left(\tfrac12\right) = 2,\ \text{and } 2\cdot\tfrac12-0-1 = 0 \;\;\checkmark\]

\[\text{Both contain the chord through } \bigl(1,2\bigr) \text{ and } \bigl(-1,6\bigr),\ \text{namely } y = -2x+4 \;\;\checkmark\]