Area enclosed by general equation of second degree

Question

What is the area of the figure enclosed by the curve: $5x^2+6xy+2y^2+7x+6y+6=0$.

My attempt: $2y^2+6(x+1)x+5x^2+7x+6=0$.

$\displaystyle y=\frac{-6(x+1)\pm\sqrt{36(x+1)^2-4(5x^2+7x+6)}}{4}$

$\displaystyle y =\frac{-3(x+1)\pm \sqrt{4x^2+11x+3}}{2}$

How do I solve it?

Answer 1

As you integrate between two curves $y_+(x)$ and $y_-(x)$, you can just consider $y_+(x)-y_-(x)$, which is (fixing your calculation errors) $$\frac{\sqrt{(6x+6)^2-4\cdot2\cdot(5x^2+7x+6)}}{2}=\sqrt{-x^2+4x-3}=\sqrt{1-(x-2)^2},$$ by completing the square.

This is just the area of a half unit circle,

$$\frac{\pi}2.$$

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For completeness,

$$I:=\int_1^3\sqrt{1-(x-2)^2}dx=\int_{-1}^1\sqrt{1-x^2}dx=\left.x\sqrt{1-x^2}\right|_{-1}^1+\int_{-1}^1\frac{x^2}{\sqrt{1-x^2}}dx \\=\int_{-1}^1\frac{1-(1-x^2)}{\sqrt{1-x^2}}dx=\left.\arcsin x\right|_{-1}^1-I$$

justifies the claim.

Answer 2

Factorize the given expression from the roots you derived and plot the equation for confirmation.

Study the graph. Recognize that the blue ellipse with axes parallel to axes has been shifted up by pushing up thin slices to form the given oblique ellipse leaving area unchanged. This follows geometrization of area/volume concepts.

Accordingly Area $ = \frac{\pi}{4} .d_1. d_2 = \frac{\pi}{4} .1 .2 = \pi/2$

Answer 3

I get that ellipse major semi-axis is $$a=\frac{3+\sqrt5}{2\sqrt2}$$ minor semi-axis is $$b=\frac{3-\sqrt5}{2\sqrt2}$$ Area is $$S=\pi ab=\pi\cdot\frac{3+\sqrt5}{2\sqrt2}\cdot\frac{3-\sqrt5}{2\sqrt2}=\frac{\pi}{2}$$