Question

Find the partial differential equation of all right circular cones whose axes
coincide with z-axis.​

Answer 1

Answer:

You need to fix the vertex of the cone, say V(a,0,0)V(a,0,0) and the semi-vertical angle of the cone, say αα.

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

The unit vector along the axis of the cone is i^i^.

Using the dot product, we can write the following equation :

VP−→−⋅i^=|VP−→−||i^|cosαVP→⋅i^=|VP→||i^|cos⁡α

i.e. (x−a)=(x−a)2+y2+z2−−−−−−−−−−−−−−−√cosα(x−a)=(x−a)2+y2+z2cos⁡α

∴,(x−a)2=[(x−a)2+y2+z2]×cos2α∴,(x−a)2=[(x−a)2+y2+z2]×cos2⁡α

That’s the required equation.

Step-by-step explanation:

Hope it helps

Answer 2

Given:

All right circular cones axes coincide with the z-axis.

To find:

The partial differential equation of all right circular cones whose axes

coincide with the z-axis.​

Solution:

$Fix the vertex of the cone, as V(0, 0,a) and the semi-vertical angle of the cone, as \alpha .Let P(x, y, z) be any point on the cone. \\The unit vector along the axis of the cone is \hat{k} . \\By dot product, we get:\\ \overrightarrow{V P} \cdot \hat{k}=|\overrightarrow{V P}||\hat{k}| \cos \alpha i.e.,\\ (z-a)=\sqrt{x^{2}+y^{2}+(z-a)^{2}} \cos \alpha \\ \therefore(z-a)^{2}=\left[x^{2}+y^{2}+(z-a)^{2}\right] \times \cos ^{2} \alpha$

Therefore, $(z-a)^{2}=\left[x^{2}+y^{2}+(z-a)^{2}\right] \times \cos ^{2} \alpha$ is the required equation.