Hello, this is Brandon, a student attending the University of Minnesota. My question is a college-level problem, and it is as follows.
Find the equation of a double cone with vertex (1,1,2) and which intersects the xy plane in a circle of radius 4. If you could please write your answer in the form R2=A(x-a)2+B(y-b)2-C(z-c)2. I am pretty sure this is the right form of the equation. The standard is z2=x2+y2. The point (a,b,c) should be the vertex and R the radius. But how do the coefficients A,B and C affect the graph?
I have had difficulty finding information on the web for variations of a double cone; your help is much appreciated.
Thank You very much.
Hi Brandon,
I am going to assume that you are looking for a circular cone given that you have the standard as z2=x2+y2. In that case the form of the equation you want is
C(z - c)2 = A(x - a)2+B(y - b)2
where (a, b, c) is the vertex of the cone. Again, since you are looking for a circular cone, for each fixed value of z, the graph is a circle so you must have A = B. Thus
C(z - 2)2 = A(x - 1)2+A(y - 1)2
A connot be zero so to simplify even further divide both sides by A to get
D(z - 2)2 = (x - 1)2+(y - 1)2
Where D = A/B.
Can you finish it now?
Penny
