Answer to Question #102910 in Analytic Geometry for ag

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

Answer: $6y^2+24zy-4yx-32y+16z^2+26-40z-4x$

The vertex is $A(1, -1, 2)$ . Let $a, b, c$ be the direction ratios of a generator of the cone. Then the equations of generator are,

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

The coordinates of any point on the generator are ($1+at$ , $-1+bt$ , $2+ct$). For some $t \in \mathbb{R}$ , ($1+at$ , $-1+bt$ , $2+ct$ ) 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,

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

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

After simplification, the required equation of the cone is,

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