## Task B-2.6.

The vertex of the parabola is at point $T(x, y), y<0$. The parabola intersects the $x$-axis at points $A(2,0)$ and $B(8,0)$. If the numerical values of the perimeter and area of triangle $ABT$ are in the ratio $4:3$, determine the equation of the parabola and the coordinates of the vertex of the parabola.
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_d4daff78382705a7c34eg-12.jpg?height=517&width=787&top_left_y=2026&top_left_x=563)

The x-coordinate of the vertex of the parabola is $x_{T}=\frac{2+8}{2}=5$.

1 point

The equation of the parabola is of the form $y=a(x-2)(x-8)$, or $y=a x^{2}-10 a x+16 a$, where $a>0$.

Therefore, the y-coordinate of the vertex of the parabola is $y_{T}=\frac{4 a c-b^{2}}{4 a}=\frac{4 a \cdot 16 a-100 a^{2}}{4 a}=-9 a$ 1 point Thus, the vertex of the parabola is the point $T(5,-9 a), a>0$.

From the ratio of the perimeter and the area of triangle $A B T$, we obtain the following sequence of equalities:

$$
\begin{gathered}
\frac{O_{\triangle A B T}}{P_{\triangle A B T}}=\frac{4}{3} \\
\frac{|A B|+|A T|+|B T|}{\frac{1}{2}|A B| \cdot\left|y_{T}\right|}=\frac{4}{3} \\
\frac{6+2 \sqrt{9+81 a^{2}}}{27 a}=\frac{4}{3}
\end{gathered}
$$

From this, after multiplying by $a$, we get the irrational equation

$$
\sqrt{9+81 a^{2}}=18 a-3
$$

which, after squaring, becomes the quadratic equation $243 a^{2}-108 a=0$ whose solutions are $a=0$ and $a=\frac{4}{9}$.

Since $a>0$, and for $a=\frac{4}{9}$ the equality

$$
\sqrt{9+81 \cdot\left(\frac{4}{9}\right)^{2}}=18 \cdot \frac{4}{9}-3
$$

is true, it follows that $a=\frac{4}{9}$, and the vertex of the parabola has coordinates $T(5,-4)$.

The equation of the given parabola is $y=\frac{4}{9}(x-2)(x-8)$.