286. Find the equations of the tangent line and the normal plane to the curve:
1) $x=t^{3}, y=t^{2}, z=t$ at the point where $t=-1$;
2) $x=y^{2}, y=z^{2}$ at the point where $z=2$.
Solution. 1) Determine the coordinates of the point of tangency: $x=-1, y=1, z=-1$ (substituting $t=-1$ into the given equations). Find the derivatives of $x, y$, and $z$ with respect to $t$ and calculate their values at the point of tangency: $\dot{x}=3 t^{2}, \dot{y}=2 t, \dot{z}=1 ; \dot{x}(-1)=3, \dot{y}(-1)=-2$, $\dot{z}(-1)=1$.

Substituting the coordinates of the point of tangency and the calculated values of the derivatives into the general equations (1) and (2), we obtain the equations of the tangent line $\frac{x+1}{3}=\frac{y-1}{-2}=\frac{2+1}{1}$ and the equation of the normal plane $3(x+1)-2(y-1)+z+1=0$ or $3 x-2 y+$ $+z+6=0$.

2) Here the curve is defined as the intersection of two surfaces. First, we transform the equations of the curve into parametric form. Setting $z=t$, we get $y=t^{2}, x=t^{4}$.*

Next, determine the coordinates of the point of tangency: $x=16, y=4$, $z=2$ and the values of the derivatives $\dot{x}, \dot{y}, \dot{z}$ at this point: $\dot{x}=4 t^{3}$, $\dot{y}=2 t, \dot{z}=1 ; \dot{x}(2)=32, \dot{y}(2)=4, \dot{z}(2)=1$.

Substituting into the general equations (1) and (2), we obtain the equations of the tangent line

$$
\frac{x-16}{32}=\frac{y-4}{4}=\frac{2--2}{1}
$$

and the equation of the normal plane

$$
32(x-16)+4(y-4)+z-2=0 \text { or } 32 x+4 y+z-530=0 .
$$