Answer to Question #141007 in Analytic Geometry for Dhruv Rawat

Question

Find the equation of the cone whose vertex is at the origin and base is the circle x=a, y^2+z^2=b^2.

Answer 1

The sphere S intersects with the plane П $\to$ x = a gives the circle

$C \to y^2+z^2=b^2$

$S \to ||p-p_0||=r$

with

$r = \frac{b}{a}\sqrt{b^2+a^2}$

$p = (x,y,z), p_0 = (x_0,y_0,z_0)$

$x_0 = a+\sqrt{r^2-b^2}, y_0=z_0=0$

Consider the line $L\to p=\lambda \vec{v}$ which passes by (0,0,0)

intersections S and L:

$||\lambda\vec{v} - p_0||=r$

$\lambda^2||\vec{v}||^2-2\lambda\langle\vec{v}, p_0\rangle+||p_0||^2=r^2$

$\lambda=\langle\vec{v}, p_0\rangle \pm \sqrt{\langle\vec{v}, p_0\rangle^2-||\vec{v}||^2(||p_0||^2-r^2)}$

the line L must be tangent to S hence

$\langle\vec{v}, p_0\rangle^2-||\vec{v}||^2(||p_0||^2-r^2)=0$

This is the locus defining the cone surface.

Putting $\vec{v}=(x,y,z)$ and applying the values for $p_0,r$ we get

$b^4x^2-a^4(y^2+z^2)-2a^2b^2(y^2+z^2)+a^2b^2(x^2+y^2+z^2)=0$