# Quadric surface S tangent to plane

Suppose that the quadric surface S is given by $$z = x^2 + x + 2y^2 + 3y$$ and the plane is given by $$x + y + z = k$$, where k is a constant.

Find the vector equation for the tangent line to the curve of intersection between S and plane when $$k=3$$ at the point (1, 0, 2).

Find the value of k for which the surface S is tangent to plane. For this plane with the value of k found, find the coordinates of the point of tangency.

For the first part of the question, I managed to derive this:

Equating S and the plane together gives - $$x^2+2x+2y^2+4y=3$$. Letting $$x=t$$, I am able to find y in terms of t, $$y=1+\sqrt{\frac{5}{2}-\frac{t^2}{2}-t}$$ and z in terms of t, $$z=2-t-\sqrt{\frac{5}{2}-\frac{t^2}{2}-t}$$.

After finding the first derivative $$f_x$$,$$f_y$$ and $$f_z$$, I am able to substitute $$x=t=1$$ into $$f_x$$,$$f_y$$ and $$f_z$$, which gives me (1, -1, 0). Hence the tangent line is : $$r(t) = (1,0,2)+t(1,-1,0)$$.

However, for the second part of the question, I am unsure of how to solve using the same method as $$k$$ is now an unknown and I am only able to derive that the normal to the plane is (1, 1, 1). May I get some help with the second part?

• If you equate $S$ and the generic plane, you will find the equation of an ellipse. For the tangency, the ellipse must degenerate to a point. – N74 Feb 9 at 17:01
• Hint: the surface normal at a point is also normal to the tangent plane at that point. – amd Feb 10 at 0:22

$$C(x,y,z) = x^2+x+2y^2+3y-z=0$$

is a paraboloid then the tangency to the plane

$$\pi(x,y,z) = x+y+z-k=0$$

has only one solution. This solution point can be found by many ways. Here we choose the following.

Substituting $$z = k - x - y$$ into $$C(x,y,z)$$ gives

$$2y^2+4y+x^2+2x-k = 0$$

Those are the points pertaining to $$C(x,y,z)\cap\pi(x,y,z)$$

Solving for $$y$$ we have

$$y = \frac 12\left(2\pm\sqrt{2}\sqrt{k+2-2x-x^2}\right)$$

At tangency we have $$\sqrt{k+2-2x-x^2}=0$$ or at

$$x = -1\pm\sqrt{k+3}$$

that should be unique hence $$k = -3$$ giving $$p_t = (-1,-1,-1)$$

The surface normal at a point is also normal to the tangent plane at that point. Rewriting the equation of the surface as $$f(x,y,z)=x^2+x+2y^2+3y-z=0$$, this condition can be written as $$\nabla f = \lambda (1,1,1)^T$$, or, since you’re working in $$\mathbb R^3$$, $$\nabla f\times(1,1,1)^T = 0$$ (which avoids introducing the extraneous variable $$\lambda$$). The latter equation expands into three linear equations, two of which are independent. From them, you can find the points on the surface at which the normal is some multiple of $$(1,1,1)$$, after which finding the corresponding value(s) of $$k$$ should be a simple task.

You can also solve this without using calculus: Working in homogeneous coordinates, if a quadric surface has the equation $$\mathbf x^TQ\mathbf x=0$$, then its tangent hyperplanes $$\mathbf\pi$$ satisfy the dual equation $$\mathbf\pi^TQ^*\mathbf\pi=0$$. For a nondegenerate quadric such as yours, the dual conic matrix $$Q^*$$ can be taken to be $$Q^{-1}$$. This gets you a simple equation in $$k$$ to solve.

Incidentally, the first part can be solved without explicitly finding the intersection curve. Since the intersection curve, whatever that might be, is planar, the tangent line must lie both on the given plane and on the tangent plane to $$S$$ at the given point. I expect that you know how to construct the tangent plane, so the problem is reduced to a straightforward linear algebra exercise of finding the intersection of a pair of planes. Here, too, the tangent plane can also be found without using calculus by taking advantage of the fact that a point on a quadric and the tangent plane at that point are a polar pair. This allows you to construct an equation for it directly from the equation of the quadric surface: in this case, the tangent plane at any point $$(x_0,y_0,z_0)$$ is $$\frac12(z+z_0)=xx_0 + \frac12(x+x_0) + 2yy_0 + \frac32(y+y_0).$$