Answer to Question #114937 in Analytic Geometry for ANJU JAYACHANDRAN

Question

Find the equation of the cone with the vertex at (1,−1,2) and the base curve as
(z+1)^2 = x+2, y = 3.

Answer 1

Solution:

The vertex $V(1,-1,2)$. Let $a,b,c$ be the direction ratios of generator of the cone.

Then the equations of the generator are

$\frac{x-1}{a}=\frac{y+1}{b}=\frac{z-2}{c}=t$ (say) (I)

The coordinates of any points on the generator are $(1+at,-1+bt,2+ct)$.

For some $t\in \R, (1+at,-1+bt,2+ct)$ lies on the guiding curve.

Therefore $((2+ct)+1)^2=(1+at)+2$ and $-1+bt=3$ . Thus, $t=\frac{4}{b}$.

From this we get

$((2+c*\frac{4}{b})+1)^2=(1+a*\frac{4}{b})+2$

From (I) we obtain:

$(3+4*\frac{z-2}{y+1})^2=3+4*\frac{x-1}{y+1}$

After simplification, the required equation of the cone is

$6y^2+24zy-4xy-32y+16z^2+26-40z-4x=0.$

Answer:

$6y^2+24zy-4xy-32y+16z^2+26-40z-4x=0.$