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

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 $$x$$:

$$E1: \ x^2=y-3$$

$$E2: \ y^2+(y-3)^2=25 \Rightarrow 2y^2-6y-16=0 \Rightarrow y=\dfrac{6\pm\sqrt{36+128}}{4}$$

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:

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

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

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