Question

Find the equation of right circular cone which has vertex at origin, axis is Y-axis & semi-vertical angle °.

Answer 1

Answer:

the equation of the right circular cone with its vertex at the origin axis along z axis and semi vertical angle a is

Answer 2

Answer:

The equation of the right circular cone which has a vertex at origin and axis is the y-axis and semi-vertical angle as $0$ is

$x^{2} +z^{2} =0$

Step-by-step explanation:

Step-1

Let P(x, y, z) be any point on the cone.

Vertex = V(0, O, O)

Step-2

Direction ratios are $x-0,y-0,z-0$

Direction cosines of the y-axis are $cos(\frac{\pi }{2}), cos(0),cos(\frac{\pi }{2})$

Direction ratios of x-axis are-

• a = $\frac{0}{\sqrt{0^{2}+0^{2}+1^{2} } }$

        a = $0$

• b = $\frac{1}{\sqrt{0^{2}+0^{2}+1^{2} } }$

        b = $1$

• c =$\frac{0}{\sqrt{0^{2}+0^{2}+1^{2} } }$

        c = $0$

Step-3

• Given the semi-vertical angle,

$\alpha =0$

$cos\alpha =cos0$

• The equation of the semi-vertical angle is given by,

$cos0 =$$\frac{0(x-0)+1(y-0)+0(z-0)}{\sqrt{1^{2}+0^{2} +0^{2}\sqrt{x^{2} +y^{2}+z^{2} }}}$

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

$\sqrt{x^{2}+y^{2}+z^{2} }=y$

Step-4

Squaring both sides of the equation-

$x^{2}+y^{2}+z^{2} = y^{2}$

Required equation= $x^{2} +z^{2}=0$