Let's determine and plot the cubic parabola that intersects the circle

$$
4 x^{2}+4 y^{2}-8 x-y-14=0
$$

at point $A(1, y0)$; the tangent of the angle between the inflectional tangent of the curve and the line $AB$ is $\frac{7}{11}$.
The coordinates of $A$ satisfy the equation of the circle. If $x=1$, then

$$
4 y^{2}-y-18=0 ; \text { from here } y=\frac{1 \pm 17}{8} . \text { Since } y<0, \quad y=-2
$$

Therefore, the coordinates of $A$ are $(1,-2)$; the coordinates of point $B$ are $(2,2)$.

The equation of the sought cubic parabola is:

$$
y=a x^{3}+b x^{2}+c x+d \ldots
$$

Here, the coefficients $a, b, c, d$ are unknown, and we need to determine them based on the conditions; thus, we need 4 equations. The parabola passes through points $A(1,-2)$ and $B(2,2)$; these points therefore satisfy the equation of the parabola, that is:

$$
\begin{array}{r}
-2=a+b+c+d \ldots \\
+2=8 a+4 b+2 c+d \ldots
\end{array}
$$

The parabola and the circle have a common tangent at $A$. The coordinates of the center $O_{1}$ of the circle are: $1, \frac{1}{8}$, that is, the abscissa of $A$ coincides with the abscissa of $O_{1}$; this implies that the tangent to the circle at $A$ is parallel to the $X$-axis, with a slope of $=0$. The radius of the circle is $2 \frac{1}{8}$.

The slope of the parabola is: $y^{\prime}=3 a x^{2}+2 b x+c$.

If $x=1$, then $y^{\prime}=0$, that is

$$
3 a+2 b+c=0 \ldots
$$

Subtracting the corresponding terms of equations (2) and (3), we get:

$$
4 a+b=4, \quad b=4(1-a) \text { and thus } c=5 a-8 \ldots
$$

Let's now determine the inflection tangent of the parabola. For the inflection point,

$$
y^{\prime \prime}=6 a x+2 b=0, \quad x_{i}=-\frac{b}{3 a}
$$

and thus

$$
y_{i}^{\prime}=3 a\left(-\frac{b}{3 a}\right)^{2}+2 b\left(-\frac{b}{3 a}\right)+c=\frac{b^{2}}{3 a}-\frac{b^{2}}{3 a}+c=c-\frac{b^{2}}{3 a}
$$

Substituting the values of $b$ and $c(3 a)$:

$$
\begin{gathered}
y_{i}^{\prime}=5 a-8-\frac{(4-4 a)^{2}}{3 a}=\frac{15 a^{2}-24 a-16+32 a-16 a^{2}}{3 a}=\frac{-a^{2}+8 a-16}{3 a} \\
y_{i}^{\prime}=-\frac{(a-4)^{2}}{3 a} \cdots
\end{gathered}
$$

The slope of the line $AB$ is: $\frac{2+2}{1}=4$.

The tangent of the angle between the inflection tangent and $AB$ is:

$$
\begin{gathered}
\frac{y_{i}^{\prime}-4}{1+4 y_{i}^{\prime}}=\frac{7}{11} \text { and thus } y_{i}^{\prime}=-3, \text { so } \\
-\frac{(a-4)^{2}}{3 a}=-3, \quad(a-4)^{2}=9 a, \quad a^{2}-17 a+16=0 \ldots
\end{gathered}
$$

Accordingly, we get two values for $a$:

$$
a_{1}=1, \quad a_{2}=16
$$

I. If $a_{1}=1$, then $b_{1}=0, c_{1}=-3$ and from (1) $d_{1}=0$.

Thus, one parabola is $y=x^{3}-3 x$.

Now $y^{\prime}=3 x^{2}-3$ and $y^{\prime \prime}=6 x$.

The inflection point is the origin.

The graph of the function can be plotted based on the following table:

| $x$ | $-\infty$ | $\ldots$ | $-\sqrt{3}$ |  | -1 |  | 0 |  | +1 |  | $\sqrt{3}$ |  | 2 | $\ldots$ | $+\infty$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $y$ | $-\infty$ | $\nearrow$ | 0 | $\nearrow$ | 2  $\max$. | $\searrow$ | 0  inf. | $\searrow$ | -2  min. | $\nearrow$ | 0 | $\nearrow$ | 2 | $\nearrow$ | $+\infty$ |
| $y^{\prime}$ | + | + | + | + | 0 | - | - | - | 0 | + | + | + | + | + | + |

![](https://cdn.mathpix.com/cropped/2024_05_02_49d45c1e5ea3b82abd8bg-2.jpg?height=515&width=480&top_left_y=434&top_left_x=817)

II. For $a_{2}=16$, we have $b_{2}=-60, c_{2}=72, d_{2}=-30$. Thus,

$$
\begin{gathered}
y=16 x^{3}-60 x^{2}+72 x-30 \\
y^{\prime}=48 x^{2}-120 x+72, \quad y^{\prime \prime}=96 x-120
\end{gathered}
$$

The graph of the function can be plotted based on the following table:

| $x$ | $-\infty$ |  | 0 |  | 1 |  | 1,25 |  | 1,5 |  | 2 | $\ldots$ | $+\infty$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $y$ | $-\infty$ | $\nearrow$ | -30 | $\nearrow$ | -2  $\max$. | $\searrow$ | $-2,5$  inf. | $\searrow$ | -3  min. | $\nearrow$ | 2 | $\nearrow$ | $+\infty$ |
| $y^{\prime}$ | + | + | + | + | 0 | - | - | - | 0 | + | + | + | + |

![](https://cdn.mathpix.com/cropped/2024_05_02_49d45c1e5ea3b82abd8bg-2.jpg?height=513&width=480&top_left_y=1489&top_left_x=844)

The function has only one zero between 1.5 and 2.

Géza Kádár (Dobó István g. VIII. o., Eger)