# Find an equation of a plane which is tangent to the graph of the paraboloid $z=x^2+4y^2+1$ and contains the origin (0, 0, 0).

Find an equation of a plane which is tangent to the graph of the paraboloid $$z=x^2+4y^2+1$$ and contains the origin (0, 0, 0).

I was able to get the partial derivative and came up with the following formula of the plane: $$2x_0(x-x_0)+8y_0(y-y_0)-z(z-z_0)=0$$

However, how do I find $$(x_0, y_0)$$, the point where the plane is tangent to the paraboloid?

• Hint: Is the origin on the paraboloid? May 25 at 8:53
• There was a typo with the equation, just fixed it. May 25 at 10:21
• Your formula for the plane is incorrect (having a $z^2$ term). May 25 at 12:55

The replies by @WindSoul and @Gaurang provide the answer to your question, but I feel like the essence is lost a bit. This reply summarises it as concisely as possible.

The equation of the tangent plane is: $$z-z_0 = 2x_0 (x-x_0) + 8y_0 (y-y_0)$$ Since $$(x_0,y_0,z_0)$$ must lie on the paraboloid, we have: $$2x_0^2 + 8y_0^2 = z_0 = x_0^2 + 4y_0^2 + 1$$ $$x_0^2 + 4y_0^2 = 1 \tag{1}$$ Since $$(0,0,0)$$ must lie on the tangent plane, we have: $$0-z_0 = 2x_0 (0-x_0) + 8y_0 (0-y)$$ $$z_0 = 2x_0^2 + 8y_0^2 = 2(x_0^2 + 4y_0^2)\tag{2}$$

Combining $$(1)$$ and $$(2)$$, we obtain $$z_0=2$$. You can freely choose $$x_0$$ and $$y_0$$, as long as $$(1)$$ holds. Examples are: $$(1,0)$$, $$(0,\frac{1}{2})$$ and $$(2^{-1/2},2^{-3/2})$$. In general, we can find $$x_0$$ and $$y_0$$ in terms of a parameter $$\theta$$: $$\begin{cases} x_0 = \cos(\theta) \\ y_0 = \frac{1}{2} \sin(\theta) \end{cases}$$ Choose any $$\theta \in [0,2\pi)$$ and you will find a valid point $$(x_0,y_0,z_0)$$ where the plane touches the paraboloid. The rest is up to you.

Step 1 Find the normal vector :

rewrite as Level surface : $$F(x,y,z)= x^2+4y^2+1-z$$

Normal vector at point $$(x_0, y_0, z_0)$$ :

$$\nabla F(x_0,y_0,z_0)= \frac{\partial F } {\partial x}~\hat{i} + \frac{\partial F } {\partial y}~\hat{j} + \frac{\partial F } {\partial z}~\hat{k}$$

$$\nabla F(x_0, y_0, z_0)= 2x_0~\hat{i} + 8y_0~\hat{j} + -1 ~\hat{k} {\tag 1 }$$

Vector on the plane :

$$(x-x_0)~ \hat{i}+(y-y_0) \hat{j} + (z-z_0) \hat{k} \tag{2}$$

$$\space$$ Normal vector (1) perpendicular to $$(2)$$ so their dot product is zero :

$$2x_0 (x-x_0) + 8y_0(y-y_0) - 1 (z-z_0) =0 \tag{3}$$

Expanding :

$$2xx_0-2x_0^2 + 8yy_0- 8y_0^2- z +z_0 =0 \tag{4}$$

With replacing $$z_0 = x_0^2+4y_0^2 +1$$ :

$$2xx_0-x_0^2-4y_0^2+8y_0y-z+1 = 0 \tag{5}$$

Since the plane contains the origin $$(x,y,z)= (0,0,0)$$

$$1 = x_0^2+4y_0^2$$

$$1 = {z_0}- 1, ~ z_0= 2$$

• The equation $z = 2x^2 + 8y^2$ describes a paraboloid, not a plane. May 25 at 12:07
• @beertje00 yes made the changes May 25 at 13:55