Interesting Integrals 11

The curve $f(x) = 5x^2+6xy+2y^2 + 7x + 6y + 6$ is a conic section. Let us check which conic section is it. Its determinant is

$\Delta = \begin{vmatrix} 5 & 3 & \frac 72 \\ 3 & 2 & 3 \\ \frac 72 & 3 & 6 \end{vmatrix} = - \frac 12 \ne 0$ .

And its discriminant $B^2-AC = 9-10 = -1 < 0$ and $A \ne C$ , therefore the curve is a ellipse.

For an ellipse of $ax^2 + 2hxy + by^2 + 2gx + 2hf + c = 0$ , the semi-axis lengths are:

$\begin{aligned} a', b' & = \sqrt{\frac{-2\Delta}{\left(h^2-ab\right)\left(\pm \sqrt{(a-b)^2+4h^2}-(a+b) \right)}} \\ & = \sqrt{\frac{1}{\left(9-10\right)\left(\pm \sqrt{9+36}-7 \right)}} \\ & = \sqrt{\frac{1}{7 \pm 3\sqrt{5}}} \\ & = \frac{\sqrt{7 \mp 3\sqrt 5}}2 \end{aligned}$

Area of ellipse $A_{ellipse} = \pi a'b' = \pi \cdot \dfrac{\sqrt{7 + 3\sqrt 5}}2 \cdot \dfrac{\sqrt{7 - 3\sqrt 5}}2 = \boxed{\dfrac \pi 2}$

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Another method to solve for the area of ellipse is to turn $f(x)$ into the form $Ax^2+Bxy+Cy^2 = 1$ .

$\begin{aligned} 5x^2+6xy+2y^2 + 7x + 6y + 6 & = 0 \\ 5(x+h)^2 + 6(x+h)(y+k) + 2(y+k)^2 + c & = 0 \\ 5(x^2 + 2hx + h^2) + 6(xy + kx + hy +kh) + 2(y^2+2ky + k^2) + c & = 0 \end{aligned}$

Equating coefficients $\begin{cases} 10h + 6k = 7 \\ 6h+4k = 6 \end{cases} \implies \begin{cases} h = -2 \\ k = \frac 92 \\ c = - \frac 12 \end{cases}$

$\begin{aligned} 5(x-2)^2 + 6(x-2)\left(y+\frac 92\right) + 2\left(y+\frac 92\right)^2 & = \frac 12 & \small \color{#3D99F6}{\text{Multiply both sides by }2} \\ 10(x-2)^2 + 12(x-2)\left(y+\frac 92\right) + 4\left(y+\frac 92\right)^2 & = 1 \end{aligned}$

The area of ellipse is given by $A_{ellipse} = \dfrac {2\pi}{\sqrt{4AC-B^2}} = \dfrac {2\pi}{\sqrt{160-144}} = \boxed{\dfrac \pi 2}$