Tangent Plane to a Paraboloid

Combining the two given equations and eliminating $z$ gives $5x^2 - 10\sqrt{2}x + 2y^2 - 10\sqrt{2}y + 20\sqrt{2}c = 0$ .

A plane tangent to a paraboloid will have one solution for $x$ and $y$ for the above equation.

To have one solution for $x$ , the discriminant of the above quadratic must be $0$ , so $(-10\sqrt{2})^2 - 4 \cdot 5 \cdot (2y^2 - 10\sqrt{2}y + 20\sqrt{2}c) = 0$ , which rearranges to $y^2 - 5\sqrt{2}y + 10\sqrt{2}c - 5 = 0$ .

To have one solution for $y$ , the discriminant of this quadratic must be $0$ , so $(-5\sqrt{2})^2 - 4 \cdot 1 \cdot (10\sqrt{2}c - 5) = 0$ , which solves to $c = \frac{7\sqrt{2}}{8} \approx \boxed{1.237}$ .

The paraboloid is defined by the following function:

$F(x,y,z) = 0.25 x^2 + 0.1 y^2 - z = 0$

The normal vector to the paraboloid at point $(x,y,z)$ is the gradient vector:

$\vec{N} = \Big(\frac{\partial{F}}{\partial{x}}, \frac{\partial{F}}{\partial{y}}, \frac{\partial{F}}{\partial{z}} \Big) = (0.5 x, 0.2 y, -1)$

By inspection, the plane normal vector is:

$\vec{N}_p = \Big( \frac{1}{2}, \frac{1}{2}, -\frac{1}{\sqrt{2}} \Big)$

At some point $(x_0, y_0, z_0)$ , the paraboloid normal vector is parallel to the plane normal vector. This is the point of tangency. Use the dot product relation:

$\vec{N} \cdot \vec{N}_p = |\vec{N}| |\vec{N}_p| \cos \theta = |\vec{N}| |\vec{N}_p|$

Expanding:

$0.25 x_0 + 0.1 y_0 + \frac{1}{\sqrt{2}} = \sqrt{0.25 x_0^2 + 0.04 y_0^2 + 1}$

Solving then yields:

$x_0 \approx 1.4142 \\ y_0 \approx 3.5355 \\ z_0 \approx 1.7500 \\ c \approx 1.2374$

The equation of the given paraboloid can be written as

$f(x, y, z) = 0.25 x^2 + 0.1 y^2 - z = 0$

The gradient of the above function of three variables is

$\nabla f = [ 0.5 x, 0.2 y, -1]^T$

At the tangency point, the gradient is along the normal of the plane, so we can write

$\nabla f = \alpha \vec{ n } = \alpha [ \frac{1}{2} , \frac{1}{2}, -\frac{1}{\sqrt{2}} ]^T$

Comparing the z-component of the above vector equation, we deduce that $\alpha = \sqrt{2}$ . And it follows that,

$x = 2 ( \frac{\sqrt{2}}{2} ) = \sqrt{2}$ and $y = 5 ( \frac{\sqrt{2}}{2} ) = \frac{5}{2} \sqrt{2}$

Since the tangency point lies on the paraboloid, then the z-coordinate is given by $z = 0.25 x^2 + 0.1 y^2 = 0.25 (2) + 0.1 (\dfrac{25}{2} ) = 0.5 + 1.25 = 1.75 = \frac{7}{4}$

Finally, since the tangency point also lies on the plane then

$c = \frac{1}{2} x + \frac{1}{2} y - \frac{1}{\sqrt{2}} z = \frac{\sqrt{2}}{2} + \frac{5 \sqrt{2}}{4} - \frac{7 \sqrt{2}}{8} = \frac{7 \sqrt{2}}{8} = 1.237$