Problems from Systems of non-linear equations

Define the equation of a parabola of the type $y = a \cdot x^{2} + b$ , with $a > 0$ and one equation of a circumference. Find its cutting points, if there are any.

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Development:

The parable and the circumference are defined, respectively, by ${\begin{array}{rcl} y & = & x^{2} + 3 \\ y^{2} + x^{2} & = & 25 \end{array}$

Before we begin to solve the system, we analyze it graphically. There is a circumference centred on the origin and a parable with the vertex at $x = 0$ . And so, the system will have:

No solution, if the vertex of the parable stays above or too far below the circumference.
A solution, if the vertex is tangent to the top point of the circumference.
Two symmetric solutions with regard to the axis if the parabola cuts the circumference at two points.

We will use the replacement method but using the square of $x$ instead of the variable $x$ :

$E 1 : x^{2} = y - 3$

$E 2 : y^{2} + (y - 3)^{2} = 25 \Rightarrow 2 y^{2} - 6 y - 16 = 0 \Rightarrow y = \frac{6 \pm \sqrt{36 + 128}}{4}$

To obtain $y > 0,$ $y_{1} = 4, 7 \Rightarrow x_{1} = 1, 3$

And, by symmetry $y_{2} = 4, 7 \Rightarrow x_{2} = - 1, 3$

Solution:

${\begin{array}{rcl} y & = & x^{2} + 3 \\ y^{2} + x^{2} & = & 25 \end{array}$

$p = (4, 7; 1, 3) q = (4, 7; - 1, 3)$

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