Question

find the equation of a right circular cone with vertex at the origin the axis of cone z axis and semi vertical angle is 30 degrees​

Answer 1

Here's the attachment

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Answer 2

Given:

Vertex at (0,0,0)

semi vertical angle=30°

To find:

find the equation of a right circular cone

Solution:

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

Direction ratios of generators are $x-0, y-0$ and $z-0$.

The direction cosines of z-axis is (0, 0, 1).

Now, to find the equation of the right circular cone, we use the formula,

$cos$ θ$=\frac{[l(x)+m(y)+n(z)]}{\sqrt{x^2 +y^2 +z^2} \sqrt{l^2+m^2+n^2} }$

We get,

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

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

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

Putting the values in the above equation, we get,

$cos$ 30°$=\frac{0(x-0)+0(y-0)+1(z-0)}{\sqrt{0^2+0^2+1^2} \sqrt{x^2+y^2+z^2} }$

⇒$\frac{\sqrt{3} }{2} =\frac{z}{\sqrt{x^2+y^2+z^2} }$

⇒$3(x^2+y^2+z^2)=4z^2$

⇒$3x^2+3y^2+3z^2-4z^2=0$

⇒$3x^2+3y^2-z^2=0$

Hence the required equation is $3x^2+3y^2-z^2=0$.