Answer:
The coordinates of the solution that lies in quadrant IV are (2, -5)
Step-by-step explanation:
We have 2 equations, the first of an ellipse and the second of a circumference.
[tex]2x^2+y^2=33\\x^2+y^2+2y=19[/tex]
To solve the system solve the second equation for x and then substitute in the first equation
[tex]x^2+y^2+2y=19\\\\x^2 = 19 -y^2 -2y[/tex]
So Substituting in the first equation we have
[tex]x^2 = 19 -y^2 -2y\\\\2(19 -y^2 -2y)+y^2=33\\\\38 -2y^2-4y +y^2 = 33\\\\-y^2-4y+5=0\\\\y^2 +4y-5 = 0[/tex]
Now we must factor the quadratic expression.
We look for two numbers that multiply as a result -5 and add them as result 4.
These numbers are -1 and 5.
Then the factors are
[tex]y^2 +4y-5 = 0\\\\(y-1)(y+5) = 0[/tex]
Therefore the system solutions are:
[tex]y = 1[/tex]; [tex]y = -5[/tex]
In the 4th quadrant the values of x are positive and the values of y are negative.
So we take the negative value of y and substitute it into the system equation to find x
[tex]y=-5\\\\2x^2+(-5)^2=33\\\\2x^2 = 33-25\\\\2x^2 = 8\\\\x^2 = 4[/tex]
[tex]x = 2[/tex], and [tex]x= -2[/tex]
In the 4th quadrant the values of x are positive
So we take the positive value of x
the coordinates of the solution that lies in quadrant IV are (2, -5)
