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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?

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

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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)$

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

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

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

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