# 1) Consider the surface x^2 + y^2/9 + z^2/9 = 1. a) Find the equation of the tangent plane to...

## Question:

1) Consider the surface {eq}x^2 + \frac{y^2}{9} + \frac{z^2}{9} = 1 {/eq}.

a) Find the equation of the tangent plane to this surface at the point {eq}(\frac{1}{3}, 2, 2) {/eq}.

b) Find a point at which the tangent plane to this surface is horizontal. Are there any other such points?

c) Find a point at which the tangent plane to this surface is vertical. Are there any other such points?

## Tangent Plane

The equation of the tangent plane to a surface {eq}f(x,y,z)=c {/eq} where {eq}c {/eq} is a constant at a point {eq}(x_0,y_0,z_0) {/eq} on the surface is given by

{eq}\begin{eqnarray*} \left.\frac{\partial z}{\partial x}\right|_{(x_0,y_0,z_0)}(x-x_0) +\left.\frac{\partial z}{\partial y}\right|_{(x_0,y_0,z_0)}(y-y_0)-(z-z_0) &=& 0. \end{eqnarray*} {/eq}

Moreover, the vector

{eq}\begin{eqnarray*} \textbf{n} &=& \left(\left.\frac{\partial z}{\partial x}\right|_{(x_0,y_0,z_0)}, \left.\frac{\partial z}{\partial x}\right|_{(x_0,y_0,z_0)},-1\right) \end{eqnarray*} {/eq}

is normal to the surface (and tangent plane) at the point {eq}(x_0,y_0,z_0) {/eq}. These formulas will be used to help answer the question. In addition, the particular surface in the question will be identified as a standard quadric surface. Once identified, parts b) and c) of the question can be answered by a geometric argument.

The question is restated. Given the surface {eq}x^2+\frac{y^2}{9}+\frac{z^2}{9}=1 {/eq} do the following.

a) Find the equation of the tangent plane to this surface at the point {eq}(\frac{1}{3},2,2) {/eq}.

b) Find a point at which the tangent plane to this surface is horizontal. Are there any other such points?

c) Find a point at which the tangent plane to this surface is vertical. Are there any other such points?

Consider part a). Find the following partial derivatives

{eq}\begin{eqnarray*} 2x+\frac{2z}{9}\cdot\frac{\partial z}{\partial x} &=& 0 \\ \frac{\partial z}{\partial x} &=& -\frac{9x}{z} \\ \frac{2y}{9}+\frac{2z}{9}\cdot\frac{\partial z}{\partial y} &=& 0 \\ \frac{\partial z}{\partial y} &=& -\frac{y}{z}. \end{eqnarray*} {/eq}

Evaluate the partial derivatives at the point {eq}(\frac{1}{3},2,2) {/eq}

{eq}\begin{eqnarray*} \left.\frac{\partial z}{\partial x}\right|_{(\frac{1}{3},2,2)} &=& -\frac{9\left(\frac{1}{3}\right)}{2} \\ &=& -\frac{3}{2} \\ \frac{\partial z}{\partial y} &=& -\frac{2}{2} \\ &=& -1. \end{eqnarray*} {/eq}

The equation of the tangent plane at the point {eq}(\frac{1}{3},2,2) {/eq} is

{eq}\begin{eqnarray*} -\frac{3}{2}\left(x-\frac{1}{3}\right)-(y-2)-(z-2) &=& 0 \\ -3\left(x-\frac{1}{3}\right)-2(y-2)-2(z-2) &=& 0 \\ -3x+1-2y+4-2z+4 &=& 0 \\ -3x-2y-2z &=& -9 \\ 3x+2y+2z &=& 9. \end{eqnarray*} {/eq}

Consider parts b) and c). Rewrite the equation of the surface in standard form

{eq}\begin{eqnarray*} \frac{(x-0)^2}{1^2}+\frac{(y-0)^2}{3^2}+\frac{(z-0)^2}{3^2} &=& 1. \end{eqnarray*} {/eq}

From the equation in standard form, we see that the surface is an ellipsoid centered at the origin.

The ellipsoid trace parallel to the {eq}xy {/eq}-plane is the ellipse

{eq}\begin{eqnarray*} \frac{(x-0)^2}{1^2}+\frac{(y-0)^2}{3^2} &=& 1. \end{eqnarray*} {/eq}

The ellipsoid trace parallel to the {eq}xz {/eq}-plane is the ellipse

{eq}\begin{eqnarray*} \frac{(x-0)^2}{1^2}+\frac{(z-0)^2}{3^2} &=& 1. \end{eqnarray*} {/eq}

The ellipsoid trace parallel to the {eq}yz {/eq}-plane is the ellipse

{eq}\begin{eqnarray*} \frac{(y-0)^2}{3^2}+\frac{(z-0)^2}{3^2} &=& 1. \end{eqnarray*} {/eq}

Consider part b). It is clear from the above observations that there are exactly two points on the ellipsoid where the tangent plane to the ellipsoid is horizontal, namely, the points {eq}(0,0,-3) {/eq} and {eq}(0,0,3) {/eq} at the bottom and top of the ellipsoid, respectively.

Consider part c). It is also clear from the above observations that at all points {eq}(x,y,0) {/eq} on the ellipsoid trace parallel to the {eq}xy {/eq}-plane, namely, all points on the ellipse

{eq}\begin{eqnarray*} \frac{(x-0)^2}{1^2}+\frac{(y-0)^2}{3^2} &=& 1 \end{eqnarray*} {/eq}

with {eq}z=0 {/eq} the tangent plane to the ellipsoid is vertical.

Parts b) and c) can also be answered analytically by taking partial derivatives. Consider part b) again. If the tangent plane is to be horizontal then the normal vector {eq}\textbf{n} {/eq} for the tangent plane must be vertical. That is, it must be of the form {eq}\textbf{n}=(0,0,c) {/eq} for some nonzero constant {eq}c {/eq}. This implies that

{eq}\begin{eqnarray*} 0 &=& \frac{\partial z}{\partial x} \\ &=& -\frac{9x}{z} \\ &=& x \\ 0 &=& \frac{\partial z}{\partial y} \\ &=& -\frac{y}{z} \\ &=& y. \end{eqnarray*} {/eq}

Since {eq}x=y=0 {/eq} then the equation of the ellipsoid reduces to

{eq}\begin{eqnarray*} \frac{z^2}{9} &=& 1 \\ z^2 &=& 9 \\ z &=& \pm3. \end{eqnarray*} {/eq}

Hence, there are two points on the ellipsoid where the tangent plane is horizontal, namely, {eq}(0,0,-3) {/eq} and {eq}(0,0,3) {/eq}. This agrees with the previous answer.

Consider part c) again. If the tangent plane is to be vertical then the normal vector {eq}\textbf{n} {/eq} for the tangent plane must be horizontal. That is, it must be of the form {eq}\textbf{n}=(a,b,0) {/eq} for some nonzero constants {eq}a {/eq} and {eq}b {/eq}. This implies that

{eq}\begin{eqnarray*} 0 &=& \frac{\partial x}{\partial z} \\ &=& -\frac{9z}{x} \\ &=& z \\ 0 &=& \frac{\partial y}{\partial z} \\ &=& -\frac{z}{y} \\ &=& z. \end{eqnarray*} {/eq}

Since {eq}z=0 {/eq} then the equation of the ellipsoid reduces to

{eq}\begin{eqnarray*} \frac{(x-0)^2}{1^2}+\frac{(y-0)^2}{3^2} &=& 1 \end{eqnarray*} {/eq}

Hence, at the points {eq}(x,y,0) {/eq} on this ellipse in the {eq}xy {/eq}-plane is where the tangent plane is vertical. This agrees with the previous answer.