Find the equation of tangent line to the intersection of the two surfaces in $P(4,-2,20)$
$$z=x^2+y^2 \,\,,\,z=2x+4y+20$$
Let $f,g:\mathbb{R}^3\rightarrow\mathbb{R}$ such that $f(x,y,z)=x^2+y^2-z\,\,\,$ and $g(x,y,z)=2x+4y-z+20$.
Then,
$$\nabla f(x,y,z)=(2x,2x,-1)\implies\nabla f(4,-2,20)=(8,-4,-1)$$
$$\nabla g(x,y,z)=(2,4,-1)\implies\nabla g(4,-2,20)=(2,4,-1)$$
The equation of tangent plane for $f$ is:
$$8x-4y-z-20=0$$
The equation of tangent plane for $g$ is:
$$2x+4y-z+20=0$$
In this step, i'm a little stuck, can someone help me?
Best Answer
The intersection of$$ z=x^2+y^2 \,\,,\,z=2x+4y+20$$ satisfies $$ (x-1)^2+(y-2)^2=25$$ The curve of intersection is parametrized by $$ x=1+5cos(t), y= 2+5sin(t) , z= (1+5cos(t))^2 +( 2+5sin(t))^2$$ Note that $$x'=-5sin(t)$$,$$ y'= 5cos(t)$$,$$z'=2(1+5cos(t))(-5sin(t))+2(2+5sin(t))(5cos(t))$$
Evaluating these derivatives at the given point $P(4,-2,20)$, we get the direction vector $V=<4,3,16>.$
The tangent line at the point $P(4,-2,20)$ is $$x=4+4t,y=-2+3t,z=20+16t$$