Q. 39 - Precalculus Enhanced with Graphing Utilities

Question

graph each equation of the system. Then solve the system to find the points of intersection.

$$\begin{gathered} 2x^2+y^2=2 \\ {x}^2-2y^2+8=0 \end{gathered}$$

Step-by-Step Solution

Answer

the point of intersection is not defined as the values are imaginary.

Step 1. Given

$$\begin{gathered} 2x^2+y^2=2 \\ {x}^2-2y^2+8=0 \end{gathered}$$

Step 2. Graphing each equation

$2x^2+y^2=2$

$x^2-2y^2+8=0$

Step 3. Calculation of intersection point

$$\begin{gathered} Consider, \\ 2x^2+y^2=2........\left(1\right) \\ {x}^2-2y^2+8=0....\left(2\right) \\ subtracting eq. 2 from eq.1\times 2 \\ {x}^2=-\frac{4}{5} \\ thus the values are imaginary. \end{gathered}$$