0
$\begingroup$

Find all tangent planes to the paraboloid $z = 100 - x^2 - y^2$ that contain the line $[x,y,z] = [10,5,40] + t[3,-1,-4]$.

I have looked around and have not been able to find a clear response for solving this type of problem.

  • 0
    Hmmm... LaTeX formatting would make this *much* more readable. Similarly, no matter the lack of a solution.. what have *you* tried and where do *you* get stuck? – 2017-01-19
  • 0
    I am new to this site and am just learning the different formatting.. sorry about that. I get stuck at where to even start because I do not know how to find the normal to the planes that would be tangent. – 2017-01-19
  • 0
    can you find tangent lines to the level curves at the point of interest? – 2017-01-19
  • 0
    I honestly do not know how. To find the tangent planes i will need the normal vector to the plane. I am confused about how the line is incorporated and how to find where on the paraboloid the tangent planes contain the line. – 2017-01-19
  • 0
    How about something basic: can you find *any* tangent plane to the surface (other than the obvious one at $(0,0,100)$, that is)? – 2017-01-19

1 Answers 1

0

$$z=f(x,y)=100-x^2-y^2$$ $$f_x(x,y)=-2x$$ $$f_y(x,y)=-2y$$

Tangent plane formula at point $P(x_0, y_0, z_0)$ is

$$z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$$

$$z-z_0=-2x_0(x-x_0)-2y_0(y-y_0)$$

$$z=-2x_0x-2y_0y+2(x_0^2+y_0^2)+z_0$$

$$z=-2x_0x-2y_0y+100+(x_0^2+y_0^2)$$

This tangent plane must contain $(10,5,4)$,

Hence $$4=-20x_0-10y_0+100+(x_0^2+y_0^2)$$

Also, let's find another point on the line, say $(10,5,4)+(3,-1,4)=(13,4,8)$

$$8=-26x_0-8y_0+100+(x_0^2+y_0^2)$$

Your task now is to recover $x_0$ and $y_0$ to recover the tangent plane $z=-2x_0x-2y_0y+100+(x_0^2+y_0^2)$