Tangent Planes to a Paraboloid

Question

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

My work so far:

Let $(a, b, c)$ rep. point of tangency (initial values).

partial derivative with respect to x = -2x = -2a

partial derivative with respect to y = -2x = -2b

$z= c - 2a(x-a)-2b(y-b)$

$z= (100-a^2-b^2) - 2a(x-a)-2b(y-b)$

$z= 100 + a^2 + b^2 - 2ax - 2by$

From here, I plugged in the given point.

$40-4t=100+a^2+b^2-2a(10+3t)-2b(5-t)$

$-60-a^2-b^2=-20a-6at-10b+2bt+4t$

Thank you!

Answer 1

Hint: Fint the normal vector $\vec n$ (the vector defining the tangent plane, perpendicular to it) of the plane.

What can you say about the value of $$\vec n \cdot [x,y,z]$$ if $[x,y,z]$ lies in the tangent plane?

Answer 2

The normal vector to the surface of the paraboloid is

$ \vec{n} = (2 x, 2 y, 1 ) $

So the equation of the tangent plane at the point $P(x_1, y_1, z_1)$ is

$( 2 x_1, 2 y_1, 1) \cdot ( x - x_1, y - y_1, z - z_1 ) = 0$

Since the given line lies in the plane, then it satisfies its equation for all values of $t$. Plug the line equation into the tangent plane equation,

$(2 x_1, 2y_1, 1) \cdot (10 + 3 t - x_1 , 5 - t - y_1, 40 - 4 t - z_1 ) = 0 $

Expand the dot product,

$ 2 x_1 (10 + 3 t) + 2 y_1 (5 - t) + (40 - 4 t) - 2 x_1^2 - 2 y_1^2 - z_1 = 0 $

But $ -2 x_1^2 - 2 y_1^2 = 2 z_1 - 200 $, therefore, the above condition on $(x_1, y_1, z_1)$ becomes

$ 2 x_1 (10 + 3 t) + 2 y_1 (5 - t) + (40 - 4 t) + z_1 - 200 = 0 $

And this can be written as

$ t ( 6 x_1 - 2 y_1 - 4 ) + (20 x_1 + 10 y_1 + z_1 - 160 ) = 0 $

Since this equation is to hold for all $t$, then we must have

$ 6 x_1 - 2 y_1 - 4 = 0 $

and

$ 20 x_1 + 10 y_1 + z_1 - 160 = 0 $

And in addition, we have that

$ z_1 = 100 - x_1^2 - y_1^2 $

Solving these three equation can be done by intersecting the first two linear equations, we will give $(x_1, y_1, z_1)$ on a straight line and then intersecting that with the third (quadratic) equation.

The solutions are

$(x_1, y_1, z_1) = (2, 4, 80) $

and

$(x_1, y_1, z_1) = (4.2 , 10.6, -30) $

Having found the two possible tangency points, then the corresponding normal vectors to the tangent planes are

$ \vec{n} = (4, 8, 1) $

and

$ \vec{n} = (8.4, 21.2, 1 ) $

Hence, the equations of the two possible tangent planes are

$ (4,8, 1) \cdot ( x - 2 , y - 4, z - 80) = 0 $

and

$ (8.4, 21.2, 1) \cdot (x - 4.2, y - 10.6, z +30) $

Simplifying, the two tangent plane equations are,

$ 4 x + 8 y + z -120 = 0 $

and

$ 84 x + 212 y + 10 z - 2300 = 0 $