Answer
The intersection of the two surfaces is a curve parametrized by two vector-valued functions:
${{\bf{r}}_1}\left( t \right) = \left( {2{t^2} - 7,t,\sqrt {9 - {t^2}} } \right)$,
${{\bf{r}}_2}\left( t \right) = \left( {2{t^2} - 7,t, - \sqrt {9 - {t^2}} } \right)$,
for $ - 3 \le t \le 3$.
Work Step by Step
We have
${y^2} - {z^2} = x - 2$
${y^2} + {z^2} = 9$
We solve the given equations for $x$ and $z$ in terms of $y$.
1. Solve for $x$
The first equation can be written $x = {y^2} - {z^2} + 2$.
From the second equation we obtain ${z^2} = 9 - {y^2}$. Thus, we can solve for $x$ by substituting ${z^2}$:
$x = {y^2} - {z^2} + 2$
$x = {y^2} - \left( {9 - {y^2}} \right) + 2$
$x = 2{y^2} - 7$
2. Solve for $z$
The second equation can be written ${z^2} = 9 - {y^2}$. So,
$z = \pm \sqrt {9 - {y^2}} $
Using $t=y$ as the parameter, we get $x = 2{t^2} - 7$ and $z = \pm \sqrt {9 - {t^2}} $. Therefore, we obtain two vector-valued functions to parametrize the entire curve, which is the intersection of the two surfaces:
${{\bf{r}}_1}\left( t \right) = \left( {2{t^2} - 7,t,\sqrt {9 - {t^2}} } \right)$,
${{\bf{r}}_2}\left( t \right) = \left( {2{t^2} - 7,t, - \sqrt {9 - {t^2}} } \right)$,
for $ - 3 \le t \le 3$.
Notice that the two signs of the square root correspond to the two halves of the curve where $z>0$ and $z<0$.
