# Cones

### Prerequisites : Coordinate Geometry I and Coordinate Geometry II

A cone is a 3D geometrical shape, formed by a set of lines connecting a common point (the vertex or the apex) to all of the points on a base, which is in a plane not containing the vertex.

Note : The last line is important. If the apex and the base are coplanar, we will get a family of straight lines.

The border of the base is known as the guiding curve of the cone and any line joining the apex and a point on the guiding curve is known as the generator.

When the guiding curve is a circle and the normal to the circle from apex passes through the center of the circle, the cone is known as right circular cone.

If the lines joining the guiding curve and apex are extended in both the directions, then the cone is known as double cone and each portion is known as a nappe.

When a plane intersects a double right circular cone, we get the conic sections, the parabola, the hyperbola and the ellipse.

• ### Equation of a Cone with Vertex at Origin

Equation of a cone, whose vertex is at origin is a homogeneous equation of 2nd degree in $x,y$ and $z$.

${ax^2 + by^2 + cz^2 + 2fyz + 2gzx + 2hxy = 0}$

When asked to find equation of such a cone, we use the 1st degree equation to make the 2nd degree equation homogeneous.

Note : The direction ratios (and the direction cosines) of a generator satisfy the equation of cone. This fact will be widely used to solve the problems on cone.

• ### Equation of a Cone passing through 3 Axes

The general equation of a cone through axes is given by

${fyz + gzx + hxy = 0}$

• ### Equation of a Right Circular Cone

Let $l,m,n$ be the direction cosines of the axis of cone and $(a,b,c)$ be the vertex. Let $\theta$ be the semi-vertical angle, which is the angle made by any line on the surface of cone and passing through the vertex.

Thus, $\hat u = l \hat i + m \hat j + n \hat k$ becomes the unit vector along the axis. Let the unit vector along the line joining $(x,y,z)$ and $(a,b,c)$ be $\hat v$. Then

${\hat v = \frac {1}{\sqrt {(x-a)^2 + (y-b)^2 + (z-c)^2}} [(x-a)\hat i + (y-b)\hat j + (z-c)\hat k]}$

The equation of cone is then given by

${\hat u \cdot \hat v = cos \theta}$

because $|\hat u|$ $= \ 1$ as well as  $|\hat v|$ $=$ $1$.

• ### General Equation of Cone and Coordinates of Vertex

${ax^2+by^2+cz^2+2fyz+2gzx+ 2gzx+2hxy+2ux+2vy+2wz+d=0}$

This is the general equation in 2nd degree. All terms except $x,y$ and $z$ are constants for a fixed shape.

Idea is, we shift the origin to a point $(\alpha, \beta, \gamma)$ so that above equation becomes homogeneous in 2nd degree.

New Coordinates : $(x = X+ \alpha, y = Y+ \beta, z = Z + \gamma)$

The single order terms are to be eliminated in order to get the homogeneous equation of 2nd degree.

• ### Enveloping Cone

Imagine a point source of light kept at $(x_1,y_1,z_1)$. Let there be a ball, whose center is at origin and radius equal to $a$. Then its equation becomes $x^2+y^2+z^2=a^2$. The rays coming out from point source are blocked by the ball. In other words, the ball forms an enveloping cone.

Let

${S = x^2 +y^2 +z^2 - a^2 , \ S_1 = x_1^2 + y_1^2 + z_1^2 -a^2}$

and

${T = xx_1 + yy_1 + zz_1 -a^2}$

Then, the equation of enveloping cone is given by

${SS_1 = T^2}$

The equation holds even if the center of the sphere is not at the origin.