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We have to find the area enclosed by the curve $$5x^2+6xy+2y^2+7x+6y+6=0.$$

I tried and I got that it is an ellipse, and I know its area is $\pi ab$ where $a$ and $b$ are the semiaxis lengths of the ellipse.

But I am unable to find the value of $a$ and $b$.

Koolman
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5 Answers5

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I think that the bests approach is to see that the general equation of an ellipse you have written above is the result of a standard ellipse that has suffered a rotation about the axes. Can you find a coordinate transformation such as

\begin{align} x' &= x \cos \theta - y \sin \theta \\ y' &= x\cos \theta + y \sin \theta \end{align} That would render the general equation in the form $$\frac{(x'-h)^{2}}{a^{2}}+ \frac{(y'-k)^{2}}{b^{2}}=m, \qquad (a, b, h, k, m) \in \mathbb{R}$$ From which you could readily find the area?

  • How to find thetha – Koolman Aug 12 '16 at 13:27
  • +1 for "suffered a rotation about the axes" but the RHS of the transformed equation will not (and should not) be zero, and the (un)rotated ellipse is not necessarily centered at the origin. –  Aug 12 '16 at 13:45
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    @tilper Quite right, I was ahead of myself, many thanks for the +1 and thanks for the picking up on my sloppy standard ellipse. –  Aug 12 '16 at 14:46
  • Why not set $m=1$? – Hrhm Aug 12 '16 at 15:06
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Hint

Let $$ax^2+bxy+cy^2+dx+ey+f=0$$ Note $$\tan 2\theta=\frac{b}{a-c}$$ or $$\cot \theta=\frac{a-c}{b}+\sqrt{1+\left(\frac{a-c}{b}\right)^2}$$ we have $$\cot \theta=\frac{1+\sqrt{5}}{2}$$ find $\sin\theta$ and $\cos\theta$ and apply \begin{cases} x=X\cos\theta-Y\sin\theta\\ y=X\sin\theta+Y\cos\theta \end{cases}

Behrouz Maleki
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The center:

find the minimum of $(5x^2 + 6xy + 2y^2 + 7x + 2y)$ $\frac {\partial}{\partial x}(5x^2 + 6xy + 2y^2 + 7x + 2y) = 10x + 6y + 7= 0\\ \frac {\partial}{\partial y}(5x^2 + 6y + 2y^2 + 7x + 2y) = 6x + 4y + 6= 0$

Solve this system of equations $x = 2, y = -4.5$

$2(x-2)^2 + 6(x-2)(y+4.5)+5(y+4.5)^2 = 0.5$ $4(x-2)^2 + 12(x-2)(y+4.5)+10(y+4.5)^2 = 1$

$\mathbf x^T\begin{bmatrix} 4&6\\ 6&10\end{bmatrix}\mathbf x =1$

Now we can find the eigenvalues of that matrix, but rather than solving for them I am just going to call them $\lambda_1,\lambda_2$

$x'^2 \lambda_1 + y'^2 \lambda_2 = 1$

$\lambda_1\lambda_2 = \frac 1{(ab)^2}$ and $\lambda_1\lambda_2$ = determinant of the matrix above = $4$.
$ab = \frac 12$

$A = \frac \pi2$

The transformed equation:

$x'^2 \lambda_1 + y'^2 \lambda_2 = 1$

$\lambda_1 = 7 + \sqrt{45}, \lambda_2 = 7 - \sqrt{45}\\ x' = (x-2) (\cos \theta) + (y+4.5)(\sin \theta)\\ y' = (x-2)(\sin \theta) - (y+4.5)(\cos \theta)\\ \theta = \frac 12 \tan^{-1} 2$

Doug M
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  • How that you have equated to 1 , is that equation of new ellipse if not then what is the equation – Koolman Aug 12 '16 at 18:12
  • How that you have equated to 0.5 , is that equation of new ellipse if not then what is the equation – Koolman Aug 15 '16 at 09:53
  • @koolman: if you substitute x'->x+2 y'->y-9/2 into your eq you get 5x'^2+2y'^2+6x'y'-1/2=0, which means that 5(x-2)^2+2(y+9/2)^2+6(x-2)(y+9/2)=1/2 (expand and see) is the same ellipse you've got. He only switched the coefficients of x^2 and y^2. – Jan-Magnus Økland Aug 15 '16 at 10:12
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For the curve $ax^2+2hxy+by^2+2gx+2fy+c = 0$, the center is given by \begin{align*} ax+hy+g &= 0\\ hx+by+f &= 0 \end{align*} and the equation of the ellipse with the axes through the center parallel to the axes is \begin{align*} ax^2+2hxy+by^2 + c' = 0 \end{align*} where $c' = gx_c + fy_c + c$, where $(x_c, y_c)$ is the center.

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The answer by Doug M solves your problem. I just want to expand on it to the general ellipse.

Assume $$ax^2+2bxy+cy^2+dx+ey+f=0$$ is an ellipse ($ac-b^2>0, a>0$). It has center $x=\frac{eb-cd}{2(ac-b^2)}, y=\frac{bd-ae}{2(ac-b^2)}$, which means that when the equation transforms to $$ax'^2+2bx'y'+cy'^2+f-\frac{ae^2-2bde+cd^2}{4(ac-b^2)}=0,$$ that is $f-\frac{ae^2-2bde+cd^2}{4(ac-b^2)}$ is negative (if not the conic is just a point or doesn't have real points). We need to multiply this by the $g>0$ so that $g(f-\frac{ae^2-2bde+cd^2}{4(ac-b^2)})=-1$. Then $agx'^2+2bgx'y'+cgy'^2-1=0$ and the area is $\frac{\pi}{g\sqrt{ac-b^2}}$ by Doug M's method.

For your equation $g=2$ and $ac-b^2=1$ so $A=\frac{\pi}{2}$.

Jan-Magnus Økland
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