find the equation of the cone whose vertex is at origin
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If the cone base is an ellipse and its vertex is at origin, then the equation is:
x²/a² + y²/b² = z² / c²
So the base elliptical surface is on x-y plane and z axis is the axis of cone.
If the base is a circle, then a = b and hence, for a right circular cone:
x² + y² = (a²/c²) z² = k z²
The general equation of a cone in three dimensions with an elliptical base and an axis inclined to x, y and z axes :
a x² + b y² + c z² + d x + e y + f z + g x y + h y z + i z x + k = 0
An equation of 2nd degree in x, y and z, if it passes through (0,0,0) then:
a x² + b y² + c z² + d x + e y + f z + g x y + h y z + i z x = 0
x²/a² + y²/b² = z² / c²
So the base elliptical surface is on x-y plane and z axis is the axis of cone.
If the base is a circle, then a = b and hence, for a right circular cone:
x² + y² = (a²/c²) z² = k z²
The general equation of a cone in three dimensions with an elliptical base and an axis inclined to x, y and z axes :
a x² + b y² + c z² + d x + e y + f z + g x y + h y z + i z x + k = 0
An equation of 2nd degree in x, y and z, if it passes through (0,0,0) then:
a x² + b y² + c z² + d x + e y + f z + g x y + h y z + i z x = 0
kvnmurty:
clik on thanks.
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1
Answer:
ax²+by²+cz²+2fyz+2gxz+2hxy=0
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