Cone

• A cone is a surface generated by a moving straight line (generator) that passes through a fixed point (vertex) and interesects a given curve (guiding curve).

• Any homogeneous equation in $x$, $y$, $z$ represents a cone with its vertex at origin.

  Note: $f(x,y,z)=0$ is homogeneous if $f(rx,ry,rz)=0$, $\mathrm{\forall }r$

• A cone is called quadric if its equation is of second degree in $x$, $y$ and $z$. The general equation of a quadric cone is $a{x}^2+b{y}^2+c{z}^2+2fyz+2gzx+2hxy=0$, with vertex as the origin.

• The equation of the cone with vertex as the origin and which passes through the curve of intersection of the plane $lx+my+nz=p$ and the surface $a{x}^2+b{y}^2+c{z}^2=1$ is given by:

• The equation of the cone with vertex as $(\alpha ,\beta ,\gamma )$ and base the conic $a{x}^2+2hxy+b{y}^2+2gx+2fy+e=0,z=0$ is given by: $a(\alpha z-yx{)}^2$ + $2h(\alpha z-yx)(\beta z-vy)$+ $b(\beta z-w{)}^2$+$2g(\alpha z-fx)(z-\gamma )$+$2f(\beta z-w)(z-\gamma )$+$c(q-z{)}^2=0$

• A right circular cone is a cone in which the generator makes a fixed angle with a fixed line passing through vertex.

  The fixed angle is caled the semi-vertical angle and the fixed line is called the axis of the cone.

• The equation of the right circular cone with vertex as origin, axis the $z-axis$ and semi-vertical angle $\alpha$ is $x^2+y^2=z^2{\tan }^2\alpha$.

• The equation of a right circular cone with vertex as the origin, axis the line ${\displaystyle \frac{x}{l}}={\displaystyle \frac{y}{m}}={\displaystyle \frac{z}{n}}$ and semi-vertical angle $\alpha$ is

• The equation of a right circular cone with vertex as $(\alpha ,\beta ,\gamma )$, semi vertical angle $\theta$ and axis having directon cosines $l$, $m$ and $n$ is given by: $[l(x-\alpha )$+$m(y-\beta )$+$n(z-\gamma ){]}^2$=$[(x-\alpha {)}^2$+$(y-\beta {)}^2$+$(z-y{)}^2]{\cos }^2\theta$

• If the direction cosines of a line satisfy a homogeneous equation, then the line is a generator of a cone.

• The equation of a cone with $x$, $y$, $z$ axes as generators and vertex as origin is given by $fyz+gzx+hxy=0$.

• The equation of the tangent (enveloping) cone from the point $P(x_1,y_1,z_1)$ to the sphere $x^2$+$y^2$+$z^2$+$2ux$+$2vy$+$2wz$+$d=0$ is given by:

where

$S=x^2+y^2+z^2+2ux+2zy+2uz+d$,

${\mathrm{S}}_1=x_1^2+y_1^2+z_1^2+2u{x}_1+2u{y}_1+2u{z}_1+d$, and

• The condition that the equation $a{x}^2+b{y}^2+c{z}^2+2fyz+2gzx+2hxy+2ux+2vy+2wz+d=0$ may represent a cone is given by: $\left|\begin{array}{cccc}a& h& g& u\\ h& b& f& v\\ g& f& c& w\\ u& v& w& d\end{array}\right|$

• Working Rule to determine the vertex of a cone represented by equation $f(x,y,z)=0$:

(i) Introduce powers of $t$ in the equation and make it homogeneous in $x$, $y$, $z$ and $t$

(ii) Find partial derivatives ${\displaystyle \frac{\mathrm{\partial }\mathrm{F}}{\mathrm{\partial }x}}$, ${\displaystyle \frac{\mathrm{\partial }\mathrm{F}}{\mathrm{\partial }y}}$, ${\displaystyle \frac{\mathrm{\partial }\mathrm{F}}{\mathrm{\partial }z}}$ and ${\displaystyle \frac{\mathrm{\partial }\mathrm{F}}{\mathrm{\partial }t}}$

(iii) Substitute $t=1$ in above four partial derivatives and equate to 0.

(iv) Solve any three equations and this solution is the vertex of the cone.

• The equation of the tangent plane at the point $(x_1,y_1,z_1)$ of the cone $a{x}^2+b{y}^2+c{z}^2+2fyz+2gzx+2hxy=0$ is given by:

• The condition that the plane $lx+my+nz=0$ may touch the cone $a{x}^2+b{y}^2+c{z}^2+2fyz+2gzx+2hxy=0$ is given by:

• The condition that the cone $a{x}^2+b{y}^2+c{z}^2+2fyz+2gzx+2hxy=0$ may have three mutually perperdicular generators is given by: $a+b+c=0$

Cone