Condition for plane $ax+by+cz+d = 0$ to touch surface $px^2+qy^2+2z=0$

Question

Q. Show that the plane $$ax+by+cz+d = 0$$ touches the surface $$px^2+qy^2+2z=0$$ if $a^2/p + b^2/q +2cd = 0$.

How to start to solve this problem?

Answer 1

Multiply the surface equation by $c$ and putting it into the equation for the plane after solving it for $cz$ gives $px^2c+qy^2c+2(-d-by-ax)=0$. Since the plane shall touch the surface, this equation shall have exactly one solution in $x$ and $y$. Its discriminant (for $x$) must be 0, which gives you $a^2+2 b c p y-c^2 p q y^2+2 c d p=0$. Now this equation (wrt to $y$) shall have exactly one solution, so the discriminant has to be 0 as well. This gives $a^2 c^2 p q+b^2 c^2 p^2+2 c^3 d p^2 q=0$. Deviding by $p^2qc^2$ yields the result. Note that here we assume that $p,q,c\neq0$.

Answer 2

Show that the plane ax+ + += 0 will touch the surface2 + 2 + 2 = 0, if 2+2+ 2 = 0. Explain in step by step anyone

Answer 3

HINT:

Eliminate $z$ to obtain the intersection projection in x-y plane as a conic

$$ {p x^2}+{q y^2}-\frac{2}{c}( ax+by+d)=0 $$

For a tangential surface contact we should have the conic and its associated intersection projection either as an ellipse of vanishing axes dimension from an ellipsoid or as a pair of parallel lines from a prismatic cylinder. Such illustrative cases for projections are plotted:

So the conic projections must be capable of being cast into quadratic forms:

$$( p x -h)^2 + (q y-k)^2 = \epsilon_1\to 0,\quad (ax+by + d +\epsilon_2)(ax+by + d + \epsilon_3)=0 \to (ax+ by +d)^2 =0 $$

where the $\epsilon \;s$ tend to zero.

Canonical form

$$f(x,y)=ax ^2 +2hxy+by ^2 +2gx+2fy+c=0.$$

has discriminant $ \Delta=h^2-ab $ which give above well known special case intersections, left as a simple further exercise. The first one has zero radius, second one has a squared straight line equation, and a hyperbola intersection projection is not admissible.

Answer 4

The normal vectors of the two surfaces are respectively $(a,b,c)$ and $(2px, 2qy, 2)$, which are aligned at the touch point, i.e. $$\frac {px}a = \frac {qy }b=\frac1c\tag 1 $$ Moreover, at the touch point $$ \frac12({p x^2}+{q y^2})=\frac{1}{c}( ax+by+d)\tag2$$ Substitute (1) into (2) to arrive at $$\frac{a^2}p+ \frac{b^2}q+2cd = 0$$

Answer 5

Using pole-and-polar relation

• Polar for a pole $(X,Y,Z)$

$$pXx+qYy+(z+Z)=0$$

• Identifying with the given plane:

$$\frac{pX}{a}=\frac{qY}{b}=\frac{1}{c}=\frac{Z}{d}$$

• For tangency, $(X,Y,Z)$ should be on the quadric, that is

\begin{align} 0 &= pX^2+qY^2+2Z \\ 0 &= p\left( \frac{a}{pc} \right)^2+ q\left( \frac{b}{qc} \right)^2+ \frac{2d}{c} \\ 0 &= \frac{a^2}{p}+\frac{b^2}{q}+2cd \end{align}