Question

The paraboloid z = 8 − x − x2 − 2y2 intersects the plane x = 3 in a parabola. Find parametric equations in terms of t for the tangent line to this parabola at the point (3, 2, −12). (Enter your answer as a comma-separated list of equations. Let x, y, and z be in terms of t.)

Answer 1

When $x=3$, we get the parabola

$z=-4-2y^2$

We can parameterize this parabola by

$\vec r(t)=(3,t,-4-2t^2)$

Then the tangent vector to this parabola is

$\vec T(t)=\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=(0,1,-4t)$

We get the point (3, 2, -12) when $t=2$, for which the tangent vector is

$\vec T(2)=(0,1,-8)$

Then the line tangent to the parabola at $t=2$ passing through the point (3, 2, -12) has vector equation

$\ell(t)=(3,2,-12)+t(0,1,-8)=(3,2+t,-12-8t)$

which in parametric form is

$\begin{cases}x(t)=3\\y(t)=2+t\\z(t)=-12-8t\end{cases}$

for $t\in\Bbb R$.

Answer 2

The parametric equation of the tangent line is $L(t)=(3,2+t,-12-8t)$

Parabola :

The equation of Paraboloid is,

                 $z =8-x-x^{2} -2y^{2}$

Equation of parabola when $x = 3$ is,

       $z=8-3-3^{2} -2y^{2} \\\\z=-4-2y^{2}$

The parametric equation of parabola will be,

     $r(t)=(3,t,-4-2t^{2} )$

Now, we have to find Tangent vector to this parabola is,

    $T(t)=\frac{dr(t)}{dt}=(0,1,-4t)$

We get, the point $(3, 2, -12)$ when $t=2$

The tangent vector will be,

 $T(2)=(0,1,-8)$

So that, the tangent line to this parabola at the point (3, 2, −12) will be,

     $L(t)=(3,2,-12)+t(0,1,-8)\\\\L(t)=(3,2+t,-12-8t)$

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