6. What is the minimum length that segment $AB$ can have if point $A$ lies on the curve $10\left(x^{2}+y^{2}\right)+60 x-80 y+249=0$, and point $B$ lies on the graph of the function $y=\frac{1}{3}|x|$? Write the square of the found length in your answer.

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Solution. The curve $10(x+3)^{2}+10(y-4)^{2}=1$ is a circle with radius $R=1 / \sqrt{10}$ and center at point $O(-3 ; 4)$. OB is the segment of the perpendicular from point $O$ to the line $\quad y=-x / 3$. Point $A$ is the intersection of $O B$ with the circle. The smallest length is the length of segment $A B$. Angle $\alpha$ is the angle of inclination of the line $y=x / 3$ to the x-axis. Then $O B=3 \cos \alpha, \operatorname{tg} \alpha=1 / 3 \Rightarrow \cos \alpha=3 / \sqrt{10}$. From this, $O B=9 / \sqrt{10} \Rightarrow A B=$

![](https://cdn.mathpix.com/cropped/2024_05_06_58f34d7e66d046d9b7d5g-20.jpg?height=463&width=548&top_left_y=1088&top_left_x=1371)
$9 / \sqrt{10}-1 / \sqrt{10}=8 / \sqrt{10} . A B^{2}=6.4$.

Answer: 6.4 .