Question

How do you find the equations for the tangent plane to the surface `x^2+2z^2=y^2` through `(1, 3, -2)`?

Answer 1

`:. x-3y-4z = 0`

Explanation:

First we rearrange the equation of the surface into the form `f(x,y,z)=0`

`x^2+2z^2 = y^2`
`:. x^2 - y^2 + 2z^2 = 0`

And so we have our function:

`f(x,y,z) = x^2 - y^2 + 2z^2`

In order to find the normal at any particular point in vector space we use the Del, or gradient operator:

`grad f(x,y,z) = (partial f)/(partial x) hat(i) + (partial f)/(partial y) hat(j) + (partial f)/(partial z) hat(k)`

remember when partially differentiating that we differentiate wrt the variable in question whilst treating the other variables as constant. And so:

`grad f = ((partial)/(partial x) (x^2 - y^2 + 2z^2))hat(i) +`
`" " ((partial)/(partial y) (x^2 - y^2 + 2z^2))hat(j) +`
`" " ((partial)/(partial z) (x^2 - y^2 + 2z^2))hat(k)`
`" "= 2xhat(i) - 2yhat(j) + 4zhat(k)`

So for the particular point `(1,3,-2)` the normal vector to the surface is given by:

`grad f(1,3,-2) = 2hat(i) -6hat(j) -8hat(k)`

So the tangent plane to the surface `x^2+2z^2 = y^2` has this normal vector and it also passes though the point `(1,3,-2)`. It will therefore have a vector equation of the form:

`vec r * vec n = vec a * vec n`

Where `vec r=((x),(y),(z))`; `vec n=( (2), (-6), (-8) )`, is the normal vector and `a` is any point in the plane

Hence, the tangent plane equation is:

`((x),(y),(z)) * ( (2), (-6),(-8) ) = ((1),(3),(-2)) * ( (2), (-6),(-8) )`
`:. (x)(2) + (y)(-6) + (z)(-2) = (1)(2) + (3)(-6) + (-2)(-8)`
`:. 2x-6y-8z = 2-18+16`
`:. 2x-6y-8z = 0`
`:. x-3y-4z = 0`

We can confirm this graphically: Here is the surface with the normal vector:

and here is the surface with the tangent plane and the normal vector: