# Find the planes tangent to the following surfaces at the indicated points.

• $x^2 + 2y^2 + 3xz = 1-$, at the point $(1, 2, \dfrac{1}{3})$
• $y^2 – x^2 = 3$, at the point (1,2,8)

This problem aims to find the 2D planes that are tangent to the given surfaces. To better understand the problem, you must be familiar with tangents, normal lines, and linear approximation techniques.

Now, tangent planes lying on a surface are planes that just brush a surface at some particular point and are also parallel to the surface at that point. One thing to note here is the point which lies on the plane. Let assume $(x_0, y_0, z_0)$ to be any point on the surface $z = f(x, y)$. If the tangent lines at $(x_0, y_0, z_0)$ to all curves on the surface departing through $(x_0, y_0, z_0)$ lie on a shared plane, that plane is known as a tangent plane to $z = f(x, y)$ at$(x_0, y_0, z_0)$.

The formula to find the tangent plane on a given smooth curved surface is:

$\nabla f(x_0). (x -x_0)=0$

Part a:

$f(x,y,z)=x^2 + 2y^2 + 3xz, x_0 = (1, 2, \dfrac{1}{3})$

Given $f(x_0)=k$:

$f(x_0)=1^2 + 2(2)^2 + 3(\dfrac{1}{3}) = 10$

$k=10$

Now calculating $\nabla f(x)$:

$\nabla f(x) = (\dfrac{d}{dx} (x^2 + 2y^2 + 3xz), \dfrac{d}{dy} (x^2 + 2y^2 + 3xz), \dfrac{d}{dz} (x^2 + 2y^2 + 3xz)$

$= (2x + 3z, 4y, 3x)$

After that, finding $\nabla f(x_0)$:

$\nabla f(1, 2, \dfrac{1}{3}) = (2 + 3 \dfrac{1}{3}, 4(2), 3)$

$\nabla f(x_0) = (3, 8, 3)$

Here, plugging the expressions in the formula:

$0=(3, 8, 3). (x-1, y-2, z – \dfrac{1}{3})$

$0=(3(x-1)+ 8(y-2) + 3(z – \dfrac{1}{3}))$

$0=(3x -3 + 8y-16 +3z – 1)$

$3x + 8y + 3z=20$

Part b:

$f(x,y,z) = y^2 – x^2, x_0=(1, 2, 8)$

$f(x_0) = 2^2 – 1^2=3$

$k=3$

Calculating $\nabla f(x)$:

$\nabla f(x)=(\dfrac{d}{dx}(y^2 – x^2) , \dfrac{d}{dy} (y^2 – x^2), \dfrac{d}{dz} (y^2 – x^2)$

$= (-2x, 2y, 0)$

After that, finding $\nabla f(x_0)$:

$\nabla f(1, 2, 8) = (-2 , 2(2), 0)$

$\nabla f(x_0) = (-2, 4, 0)$

Again, plugging the expressions in the formula:

$0 = (-2, 4, 0). (x-1, y-2, z – 8) = -2(x-1)+ 4(y-2) + 0(z – 8)$

$0 = (-2x +2 + 4y-8)$

$2y-x = 3$

Part a: $3x + 8y + 3z = 20$ is the plane tangent to the surface $x^2 + 2y^2 +3xz =1$ at the point $(1,2,\dfrac{1}{3})$.

Part b: $2y-x = 3$ is the plane tangent to the surface $y^2 -x^2 = 3$ at the point $(1,2,8)$.

## Example

Find the plane tangent to the given surface at the indicated point. $xyz = 1$, at the point $(1,1,1)$.

$f(x,y,z) = (xyz), x_0 = (1, 1, 1)$

$f(x_0) = k = 1$

Now calculating $\nabla f(x)$:

$\nabla f(x) = (\dfrac{d}{dx}(xyz) , \dfrac{d}{dy} (xyz), \dfrac{d}{dz} (xyz)$

$= (yz, xz, xy)$

After that, finding $\nabla f(x_0)$:

$\nabla f(1, 1, 1) = (1 , 1, 1)$

$\nabla f(x_0) = (1, 1, 1)$

Here, plugging the expressions in the formula:

$0 = (1, 1, 1). (x-1, y-1, z – 1) = 1(x-1)+ 1(y-1) + 1(z – 1)$

\[x+y+z=3\