how do I find the equation of the tangent plane to the parametric surface?

my professor didn't cover this material that well so I'm not sure how to do this one. the question is: Find an equation of the tangent plane to the parametric surface $x=2r\cos\theta$, $y=5r\sin\theta$, $z=r$, at the point $(2\sqrt2$, $5\sqrt2$, 2) where $r = 2$ and $\theta = \pi/4$

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Example 3 here might help. – Dylan Moreland Dec 18 '11 at 19:17
@Dylan I love his site; overall, the best student-centered, first/second year math courses site I've encountered on the web. – David Mitra Dec 18 '11 at 19:19

The parametric surface has the following parametric equation: $$f(r,\theta)=(x,y,z)=(2r\cos\theta,5r\sin\theta,r).$$ Then we have $$f_r(2,\pi/4)=(2\cos(\pi/4),5\sin(\pi/4),1)=(\sqrt{2},5\sqrt{2}/2,1),$$ $$f_\theta(2,\pi/4)=(-2\cdot 2\cdot\sin(\pi/4),5\cdot 2\cdot\cos(\pi/4),0)=(-2\sqrt{2},5\sqrt{2},0).$$ Therefore, the unit normal $n$ of the tangent plane at $(r,\theta)=(2,\pi/4)$ is equal to $$n=\frac{f_r(2,\pi/4)\times f_\theta(2,\pi/4)}{\|f_r(2,\pi/4)\times f_\theta(2,\pi/4)\|}=\frac{1}{\sqrt{458}}(-5\sqrt{2},-2\sqrt{2},20).$$ Since the tangent plane at $(r,\theta)=(2,\pi/4)$ passes through the point $f(2,\pi/4)=(2\sqrt{2},5\sqrt{2},2)$, the equation of the tangent plane is given by $n\cdot(x-2\sqrt{2},y-5\sqrt{2},z-2)=0$, or equivalently, $$(-5\sqrt{2},-2\sqrt{2},20)\cdot(x-2\sqrt{2},y-5\sqrt{2},z-2)=0,$$ that is $$5\sqrt{2}x+2\sqrt{2}y-20z=0.$$