6. What is the minimum distance that can be between two points, one of which lies on the graph of the function $y=x^{2}$, and the other - on the curve defined by the equation $4 x^{2}+4 y^{2}-48 x-24 y+163=0 . \quad$ In your answer, write the square of the found distance. (12 points)

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# Solution.

Transform the equation of the second curve by completing the squares: $(x-3)^{2}+(y-6)^{2}=17 / 4$. The second curve is a circle with center at point $(6 ; 3)$ and radius $\sqrt{17} / 2$. We need to find the minimum distance from the center of this circle to a point lying on the parabola $\quad y=x^{2} . \quad$ The square of the distance from the point ( $6 ; 3$ ) to the point $\left(x ; x^{2}\right)$ is $h(x)=(x-6)^{2}+$ $\left(x^{2}-3\right)^{2}$. Let's find the derivative of the function $h(x)$:

![](https://cdn.mathpix.com/cropped/2024_05_06_555865d131a60f67fb06g-4.jpg?height=785&width=1017&top_left_y=1686&top_left_x=908)
$h'(x)=2(x-6)+4x(x^2-3)=4x^3-10x-12=(x-2)(4x^2+8x+6)=0$, At the point $x=2$, the derivative is zero and changes sign from negative to positive, therefore, it is a minimum point of the function $h(x), h_{\min }=17$. The minimum
distance between points on the parabola and on the circle is then $\sqrt{17} / 2$, the square of the distance is $17 / 4$ or 4.25.

## Answer: 4.25