Help finding the equation of a cone in space

Question

The line

$\frac{x-2}{2}=\frac{y+1}{-2}=\frac{z+1}{-1}$

Is the axis of a circular cone with vertex on the xy-plane. Find the equation of the cone, if the point $M_1(1,1,-\frac{5}{2})$ is on the surface.

Thus far, I have found that the vertex is the point $V(0,1,0)$, and that the radius of the cone at the point $M_1$ is $\sqrt{5}$.

Answer 1

Hint:

The axis of the cone is in the direction $(2,-2,-1)$. If $P(x,y,z)$ is a point on the cone, then $\vec{VP}$ and $\vec{VM}$ make the same angle with $(2,-2,-1)$.

Use dot product.

Answer 2

Given the cone axis by the line

$$ L_1\to p = p_0 + \lambda\vec v $$

with $p = (x,y,z),\ \ p_0 = (2,-1,-1),\ \ \vec v = (2,-2,-1)$

the sphere $\Sigma\to ||p-O|| = r$ such that $O\in L_1$ and $M_1\in\Sigma$ is determined solving for $O,\lambda$.

$$ (V-M_1)\cdot(O-M_1) = 0\\ O = V + \lambda\vec v $$

obtaining

$$ O = \frac 19(29,-20,-29) $$

and

$$ r = ||M_1-O|| $$

Once we have determined $\Sigma$ the next step is to determine the locus for the tangent lines to $\Sigma$ passing by $V$

Now giving a generic line passing by $V$ as

$$ L_2\to p = V + \mu \vec \omega $$

$\Sigma\cap L_2$ is determined as follows:

$$ ||V+\mu\vec\omega-O||^2 = r^2 $$

or

$$ ||V-O||^2+2\mu(V-O)\cdot\vec\omega +\mu^2||\vec\omega||^2 = r^2 $$

an the intersection points are associated to the $\mu$ values

$$ \mu = \frac{1}{2||\vec\omega||^2}\left(2(V-O)\cdot\vec\omega\pm\sqrt{4((V-O)\cdot\vec\omega)^2-4||\vec\omega||^2(||V-O||^2-r^2)}\right) $$

but at tangency we have

$$ ((V-O)\cdot\vec\omega)^2-||\vec\omega||^2(||V-O||^2-r^2) = 0 $$

or

$$ ((V-O)\cdot(p-V))^2-||p-V||^2(||V-O||^2-r^2) = 0 $$

with $p = (x,y,z)$ which is the cone equation.

hence

$$ \frac{841}{324} (-2 x+2 y+z-2)^2-\frac{29}{4} \left(x^2+(y-1)^2+z^2\right) = 0 $$

In red $O$ in black $M_1$ and in green $V$

Answer 3

The cone with the vertex $V$ and axis defined by vector $\vec{a}$ can be defined as the set of points $P$ such that vector $\vec{VP}$ forms a given angle with line of vector $a$, that is $$ \frac{\vec{VP}\cdot\vec{a}}{|VP||a|} = \cos \alpha = {\rm const.}$$ Just for the note, a double cone would have equation $$ \frac{(\vec{VP}\cdot\vec{a})^2}{|VP|^2|a|^2} = \cos^2 \alpha = {\rm const.}$$

You have $V = (0,1,0)$, you can find $\vec a \sim (2,-2,-1)$, and you know that $M = (1,1,-\frac52)$ satisfies this equation. Therefore you can find $$ \cos \alpha = \frac{\vec{VM}\cdot\vec{a}}{|VM||a|} = \frac{(1,0,-\frac52)\cdot(2,-2,-1)}{\sqrt{\frac{29}{4}}\sqrt{9}} = \frac{3}{\sqrt{29}}$$

Therefore points $P=(x,y,z)$ of the cone need to satisfy the equation $$ \frac{(x,y-1,z)\cdot(2,-2,-1)}{\sqrt{x^2+(y-1)^2+z^2}\sqrt{9}}= \frac{3}{\sqrt{29}}$$ that is $$ \frac{2x-2y-z+2}{\sqrt{x^2+(y-1)^2+z^2}} = \frac{9}{\sqrt{29}}$$