Answer to Question #121531 in Analytic Geometry for Zian

Question

Find a centre and the principal axes of the following conicoid. Is it ellip-
soids, hyperboloids or something else?
3x2 + 5y2 + 3z2 + 2yz + 2zx + 2xy - 4x - 8z + 5 = 0

Answer 1

Let,

$$f(x,y,x)=3x^2 + 5y^2 + 3z^2 + 2yz + 2zx + 2xy - 4x - 8z + 5 = 0$$

Thus, associated quadratic form is

$$Q(x,y,z)=3x^2 + 5y^2 + 3z^2 + 2yz + 2zx + 2xy$$

Thus, matrix associated with quadratics form is $A=\begin{bmatrix} 3 & 1&1\\ 1 & 5&1\\ 1&1&3 \end{bmatrix}$

Note that, the eigenvalues of A is

$$\begin{vmatrix} 3-\lambda & 1&1\\ 1 & 5-\lambda&1\\ 1&1&3-\lambda \end{vmatrix}=0\implies (\lambda-2)(\lambda-3)(\lambda-6)=0$$

$(\lambda_1,\lambda_2,\lambda_3)=(2,3,6)$ in which all are positive,thus A is positive definite ,thus Hessian matrix $H=2A$ is also positive definite.Hence $f(x,y,z)$ has one critical point and which is center of the conic and can be found by

$$Hx=-p$$

where, $p=\begin{bmatrix} -4 \\ 0\\-8 \end{bmatrix}$ $\implies x=-H^{-1}p$ ,On plugin the values we get, center

$$x=(\frac{1}{3},-\frac{1}{3},\frac{4}{3})^T$$

Now, corresponding to each eigenvalues of A , eigen vectors is

$$Av_1=\lambda_1v_1\\ Av_2=\lambda_1v_2\\ Av_3=\lambda_1v_3\\$$

On solving we get,

$$v_1=(-1,0,1)^T\\ v_2=(1,-1,1)^T\\ v_3=(1,2,1)^T$$

Thus ,respective unit vector is,

$$\hat{v_1}=\frac{1}{\sqrt{2}}(-1,0,1)^T\\ \hat{v_2}=\frac{1}{\sqrt{3}}(1,-1,1)^T\\ \hat{v_3}=\frac{1}{\sqrt{6}}(1,2,1)^T$$

Hence, principle axis, $l_1,l_2,l_3$ will be

$$l_1=\{(\frac{1}{3},-\frac{1}{3},\frac{4}{3})^T+t\frac{1}{\sqrt{2}}(-1,0,1)^T|t\in \mathbb{R}\}\\ l_2=\{(\frac{1}{3},-\frac{1}{3},\frac{4}{3})^T+t\frac{1}{\sqrt{3}}(1,-1,1)^T|t\in \mathbb{R}\}\\ l_3=\{(\frac{1}{3},-\frac{1}{3},\frac{4}{3})^T+t\frac{1}{\sqrt{6}}(1,2,1)^T|t\in \mathbb{R}\}\\$$

Moreover, consider the orthogonal matrix

$$\Gamma=(v_1,v_2,v_3)$$

And, orthogonal transformation

$$x'=\Gamma^Tx$$

where,$x=(x_1,y_1,z_1)\&x'=(x'_1,y'_1,z'_z)$

Thus,

$$Q(x')=x'^T\Gamma^TA\Gamma x=2x'^2+3y'^2+6z'^2$$

Let,

$$P(x,y,z)=- 4x - 8z \&r=5$$

Thus,

$$P(x')=P(\Gamma x)$$

Hence, $f(x)$ transforms to $Q(x')+P(x')+r=0$ .

Now, on simplification we get,

$$2X^2+3Y^2+6Z^2=c$$

Where,

$$X=x'-\frac{1}{\sqrt{2}}\\ Y=y'-\frac{2}{\sqrt{3}}\\ Z=z'-\frac{1}{\sqrt{6}}\\$$

and $c>0$ a constant.

Therefore, given conicoid is an ellipsoid