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.
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:
$$\left\{ \begin{array} {rcl} y & = & x^2+3 \\ y^2+x^2 &=& 25 \end{array}\right.$$
$$p=(4,7;1,3) \\ q=(4,7;-1,3)$$