The statement says:
Given the surface $y = A - 2x^2 + Bz^2$ find $A$ and $B$ knowing that the interesection of the surface with the plane $y = 3$ is the curve whose parametric equations are $(x, y, z) = (2\sin{(t)}, 3, 2\cos{(t)}),\;0\leq t<2\pi$. Get intersection with coordinate planes and graph.
I tried this:
Let's call $\sigma$ to the surface $$\begin{array}{clc} &-2x^2 + Bz^2+A-y=0&(1)\\ \Rightarrow& \sigma\;\cap\;(y=3):-2x^2+Bz^2+A-3=0&\\ \Rightarrow& -2x^2+Bz^2=3-A&\\ \Rightarrow& \dfrac{x^2}{-\frac{3-A}{2}}+\dfrac{z^2}{\frac{3-A}{B}}=1.& \end{array}$$
Now, $C$ can be written as $$\begin{array}{cl} &C:\begin{cases}x&=&2\sin{(t)}\\y&=&3\\z&=&2\cos{(t)}\end{cases},0\leq t<2\pi\\ \Rightarrow&{\left(\dfrac x2\right)}^2+{\left(\dfrac z2\right)}^2=1\\ \Rightarrow&\dfrac{x^2}{4}+\dfrac{z^2}{4}=1. \end{array}$$
Equating both equations $$\begin{array}{cl} &\sigma=C\\ \Rightarrow&\begin{cases}-\dfrac{3-A}{2}&=&4\\\dfrac{3-A}{B}&=&4\end{cases}\\ \Rightarrow&A=11\;\wedge\;B=-2\\ \end{array}$$
Thus from $(1)$ we get $$\sigma:-2x^2-y-2z^2=-11\\\boxed{\sigma:\dfrac{x^2}{\frac {11}2}+\dfrac{y}{11}+\dfrac{z^2}{\frac{11}2}=1},$$ which represents an infinite paraboloid.
The intersections with the coordinate planes are:
- Intersection with the $xy$-plane: $\dfrac{x^2}{\frac{11}2}+\dfrac{y}{11}=1\Rightarrow y=-2x^2+11\qquad(2)$
- Intersection with the $yz$-plane: $\dfrac{y}{11}+\dfrac{z^2}{\frac{11}2}=1\Rightarrow y=-2z^2+11\qquad(3)$
- Intersection with the $xz$-plane: $\dfrac{x^2}{\frac {11}2}+\dfrac{z^2}{\frac{11}2}=1\qquad\qquad\qquad\qquad\qquad(4)$
Are my reasoning correct?
Thanks!