Find the equation of tangent line to the intersection of the two surfaces in $P(4,-2,20)$ $z=x^2+y^2 \,\,,\,z=2x+4y+20$

Question

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?

Answer 1

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 $$x=4+4t,y=-2+3t,z=20+16t$$

Answer 2

As have been answered in the comments to your question: since you have computed the tangent planes to both of the surfaces you are already done. If you do not see this then simply recall that a line in the 3-dim space is given by two linear equations in 3 variables. Each of the equation describes a plane and the intersection of the planes defines a line. In your case the planes are the tangent planes and thus their interesection is the tangent line to the intersection of the surfaces. This line is determined be the set of equations uniquelly (if the equations are linearly independent and have a non-empty solution).

As far as the point $P = (-4,2,20)$ is considered, only thing you can do with this guy is to check whether $P$ lies on the tangent line or not. So, there are only two options: 1) $P$ lies on the tangent line, or, 2) $P$ does not lie on the tangent line. Since your line is already determined, these two options corresponds to two possible answers: 1) the tangent line you are looking for is described by the system you found, i.e. $$ 8x - 4y - z - 20 = 0 \\ 2x + 4y - z + 20 = 0 $$ 2) the tangent line you are looking for does not exist.

Let us ponder a bit about these two options. Since you have computed the tangent planes using the point $P$ it would be very strange if the intersections of the planes would not contain $P$ (this should be clear). Indeed, you can check by direct computation (simply input the coordinates of $P$ in the equations describing the planes) that $P$ lies on your line. Hence we are done.

If you are looking for parametric expression of your line then (as was described in the comments as well) you can do the following. Add the first equation of the tangent planes to the second one to get $$ 5x = z \ .$$ Insert this result in either of the equations, say the second one, to get $$y = 5 - \frac{7}{4}x \ .$$ Now you can describe your line as the set of all points in $\mathbb{R}^3$ that are of the form $(x,5-\frac{7}{4}x, 5x)$. In other words, if we denote your line by $l$ then we can write $$ l = \{ [0,5,0] + x (1, -\frac{7}{4}, 5) | \ x \in \mathbb{R} \} \ ,$$ which is a parametric description of the line. Again, by direct computation you can check that $P$ lies in $l$.

Let me comment a bit on the other approach you could take. Instead of computing firstly the tangent planes of both surfaces and then taking their interesction, you could start with inserting one surface equation in the other to get a non-linear equation of two variables. It would be some sort of ellipse (since $z = x^2 + y^2$ is a paraboloid and we take intersection with $z = 2x + 4y + 20$ which is a plane). Then you would compute the tangent line to this curve.

Answer 3

To continue from this point, simply solve the system of linear equations that you’ve got using your favorite method. For example, Gaussian elimination produces the solution $(0,-5,0)+t(4,3,20)$.

Working out the two tangent plane equations explicitly isn’t really necessary to solve this problem, though. Being the intersection of these two planes, the tangent line is perpendicular to both planes’ normals, so you can find the direction vector of the line by taking their cross product. You already know a point that must be on the line, so with that you can immediately write down a parametric equation for the line: $\nabla f\times\nabla g = (8,-4,-1)\times(2,4,-1) = (8,6,40)$, so the tangent line is $(4,-2,20)+\lambda(8,6,40)$. A simple check verifies that this is the same line as the one at top. The implicit equivalent of this is $${x-4\over4}={y+2\over3}={z-20\over20}.$$

Note, by the way, that since the second surface is a plane, it is everywhere its own tangent plane.