What does it mean to solve the equation of an ellipse as a quadratic?

Question

An illustration required me to find out the area of the curve $$5x^2 + 6xy + 2y^2 + 7x + +6y + 6 = 0 $$

They proceeded to solve the equation as a quadratic obtaining $y_1$ and $y_2$ as the two branches.

Why are $y_1$ and $y_2$ as depicted in the diagram?

^{if it isn't too broad what does it mean to solve a quadratic equation for any conic section (parabola, ellipse, hyperbola, circle.}

Answer 1

The point of this is to write $y$ as a function of $x$ - well, actually, two functions. So, then, we treat $x$ as a constant and sort everything by powers of $y$: \begin{align*} 0 &= 5x^2+6xy+2y^2+7x+6y+6\\ 0 &= 2y^2 + (6xy+6y) + (5x^2+7x+6)\\ 0 &= 2y^2 + (6x+6)y + (5x^2+7x+6)\end{align*} Since we're treating $x$ as a constant, that's just a quadratic equation in $y$ - and we know how to solve quadratic equations: \begin{align*}y &= \frac{-(6x+6)\pm\sqrt{(6x+6)^2-4\cdot 2\cdot (5x^2+7x+6)}}{2\cdot 2}\\ &= \frac{-6(x+1)\pm\sqrt{36x^2+72x+36-40x^2-56x-48}}{4}\\ &= \frac{-6(x+1)\pm\sqrt{-4x^2+16x-12}}{4}\\ y &= \frac{-3(x+1)\pm\sqrt{(x-1)(3-x)}}{2}\end{align*} There it is. That quadratic equation has a $\pm$ sign in it, and both choices for the square root are viable. If we take the $+$ sign, we get a larger value of $y$, for the top curve $y_1$ drawn in pink. If we take the $-$ sign, we get a smaller value of $y$, for the bottom curve $y_2$ drawn in green.

Similar reasoning applies to any conic section; if we treat one variable as a constant, the equation becomes a linear or quadratic equation in the remaining variable, which we can solve explicitly.

Answer 2

I get $$y_1=-\frac{\sqrt{-{{x}^{2}}+4 x-3}+3 x+3}{2}$$ $$y_2=\frac{\sqrt{-{{x}^{2}}+4 x-3}-3 x-3}{2}$$ Area of the ellipse is $$S=\int_1^3(y_2-y_1)dx\\=\int_1^3\sqrt{-x^2+4x-3}dx=\frac{\pi}{2}$$