Show that this section of the cone is a hyperbola

Question

Show that section of a cone, with vertex at origin and base $x=a$ & $y^2+z^2=b^2$, intersected by a plane parallel to $XY$ axes is a hyperbola.

Answer 1

Equation of the cone:

$$ \dfrac{y^2 + z^2}{b^2} = \dfrac{x^2}{a^2} $$

Equation of the plane:

$$ z= c $$

Substitute $z=c$ in the equation of the cone

$$ \dfrac{y^2 + c^2}{b^2} = \dfrac{x^2}{a^2} $$ or, $$ \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = \dfrac{c^2}{b^2}$$

which is nothing but a hyperbola.

Answer 2

Allow that at any point A along the x axis, the cross-sectional radius is RA=B, where R is constant.

Then for any plane of fixed Z:

y^2 + Z^2 = (RA)^2

y^2 = R^2 x^2 - Z^2

which is the same as y^2 = Cx^2-D, where C=R^2 and D=Z^2 are constants, which is the formula for a hyperbola.

Answer 3

The equation should be $y^2+z^2=x^2$ instead of what you wrote (otherwise it's not a cone). If $z=c$ is constant then the equation gives $y^2+c^2=x^2$ or $y^2-x^2=-c^2$, which is an equation of a hyperbola. (Delete the equation $x=a$.)