$\begingroup$If a center of an ellipsoid $E$ is $(1,k,0)$ and the trace of $E$ when $z=0$ is $x^2+y^2-2x-8y+16=0$ and when $x=1$ it is $-9y^2-z^2+72y=135$. Determine the equation of E
So, the ellipsoid has the form $\frac{(x-1)^2}{a^2}+\frac{(y-k)^2}{b^2}+\frac{z^2}{c^2}=1$
So, when $z=0$ $$\frac{(x-1)^2}{a^2}+\frac{(y-k)^2}{b^2}-1=x^2+y^2-2x-8y+16=0$$ And when $x=1$ $$\frac{(y-k)^2}{b^2}+\frac{z^2}{c^2}-1=-9y^2-z^2+72y-135=0$$
Am I doing it right? Is it just a matter of finding $a,b,c,k$ with those two equations?
$\endgroup$1 Answer
$\begingroup$The trace of $E$ when $z=0$ is $x^2+y^2-2x-8y+16=0$ which is equivalent to $$C_1(x^2+y^2-2x-8y+16)=0$$ $C_1$ is an arbitrary constant$\neq 0$.
The same for $-9y^2-z^2+72y=135$ which is equivalent to $$C_2(-9y^2-z^2+72y-135)=0$$ $C_2$ is an arbitrary constant$\neq 0$.
The mistake was to set $C_1=C_2=1$ which isn't correct.
So, when $z=0$ $$\frac{(x-1)^2}{a^2}+\frac{(y-k)^2}{b^2}-1=C_1(x^2+y^2-2x-8y+16)=0$$ And when $x=1$ $$\frac{(y-k)^2}{b^2}+\frac{z^2}{c^2}-1=C_2(-9y^2-z^2+72y-135)=0$$
You will get : $\quad C_1=\frac{1}{a^2}\quad;\quad C_2=-\frac{1}{c^2}\quad;\quad k=4 \quad;\quad a=\pm 1 \quad;\quad b=\pm 1 \quad;\quad c=\pm 3 .$
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