Answer: (x,y) = (-4, 5) and (x,y) = (2/3, -13/3)
Explanation
Let's isolate y in the 2nd equation.
2x + y = -3
y = -2x-3
Then plug this into the 1st equation.
2y^2 + x^2 = -6x + 42
2(-2x-3)^2 + x^2 = -6x + 42
2(4x^2+12x+9) + x^2 = -6x+42
8x^2+24x+18+x^2 = -6x+42
9x^2+24x+18+6x-42 = 0
9x^2+30x-24 = 0
This equation is of the form ax^2+bx+c = 0, so we can use the quadratic formula to solve.
Plug in a = 9, b = 30, c = -24 into the quadratic formula.
[tex]\text{x} = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\text{x} = \frac{-30\pm\sqrt{(30)^2-4(9)(-24)}}{2(9)}\\\\\text{x} = \frac{-30\pm\sqrt{900 + 864}}{18}\\\\\text{x} = \frac{-30\pm\sqrt{1764}}{18}\\\\\text{x} = \frac{-30\pm42}{18}\\\\\text{x} = \frac{-30+42}{18} \ \text{ or } \ \text{x} = \frac{-30-42}{18}\\\\\text{x} = \frac{12}{18} \ \text{ or } \ \text{x} = \frac{-72}{18}\\\\\text{x} = \frac{2}{3} \ \text{ or } \ \text{x} = -4\\\\[/tex]
If x = 2/3, then,
y = -2x-3
y = -2(2/3)-3
y = -4/3 - 3
y = -4/3 - 9/3
y = -13/3
Therefore, one of the points of intersection between the line and the ellipse is (x,y) = (2/3, -13/3)
The other point of intersection is (x,y) = (-4, 5) as shown in this scratch work below.
y = -2x-3
y = -2(-4)-3
y = 8-3
y = 5
I confirmed each answer using GeoGebra.
