Question

Sa se determine numarul punctelor de intersectie dintre parabolele y^2=4x si x^2=4y

Answer 1

[tex]\left\{ \begin{array}{c} y^2=4x \\ x^2 = 4y \end{array} \right \Rightarrow \left\{ \begin{array}{c} y^2-4x=0 \\ x^2-4y= 0 \end{array} \right \Rightarrow \left\{ \begin{array}{c} 4x = y^2 \\ x^2-4y = 0 \end{array} \right \Rightarrow \\ \\ [/tex]

$\Rightarrow \left\{ \begin{array}{c} x = \dfrac{y^2}{4} \\ \Big(\dfrac{y^2}{4}\Big)^2 -4y= 0 \end{array} \right \Rightarrow \left\{ \begin{array}{c} x = \dfrac{y^2}{4} \\ \dfrac{y^4}{16}} -4y= 0 \end{array} \right \Rightarrow \left\{ \begin{array}{c} x = \dfrac{y^2}{4} \\ y^4-4\cdot 16y= 0 \end{array} \right \Rightarrow \\ \\ \Rightarrow \left\{ \begin{array}{c} x = \dfrac{y^2}{4} \\ y^4-64y= 0 \end{array} \right \Rightarrow \left\{ \begin{array}{c} x = \dfrac{y^2}{4} \\ y(y^3-64)= 0 \end{array} \right |$

$\boxed{1} \quad y = 0 \Rightarrow x = \dfrac{0^2}4} \Rightarrow x = 0 \Rightarrow \boxed{(x,y) = (0,0)} \\ \\ \boxed{2} \quad y^3-64 = 0 \Rightarrow y^3-4^3 = 0 \Rightarrow (y-4)(y^2+4y+4^2) = 0 \\ \\ \boxed{2_1} \quad y-4 = 0 \Rightarrow y = 4 \Rightarrow x = \dfrac{4^2}{4} \Rightarrow x = 4 \Rightarrow \boxed{(x,y) = (4,4)} \\ \boxed{2_2} \quad y^2+4y+4^2 = 0 \Rightarrow y^2+4y+16 = 0 \\ \Delta = 16 - 64 = -48 \ \textless \ 0 \Rightarrow y \notin\mathbb_{R}$


$\\ Din \boxed{1} , \boxed{2} \Rightarrow (x,y) = \Big\{(0,0); (4,4)\Big\} \\ \\ Avem 2 solutii ale sistemului, deci, avem, 2 puncte de intersectie. \\ \\ \Rightarrow \boxed{\boxed{S = \big\{2\big\}}}$