Question

If x/1=y/1=z/-1 represents one of the three mutually perpendicular generators of the cone 3xy+8xz-5yz=0, find the equations of the other two.

Answer 1

Answer on Question #78623 – Math – Analytic Geometry

Question

If $x/1 = y/1 = z/-1$ represents one of the three mutually perpendicular generators of the cone $3xy + 8xz - 5yz = 0$, find the equations of the other two.

Solution

Cone:

$$C \rightarrow a x ^ {2} + b y ^ {2} + c z ^ {2} + 2 f y z + 2 g z x + 2 h x y = 0$$

One of its generators:

$$L _ {1} \rightarrow \frac {x}{l} = \frac {y}{m} = \frac {z}{n}$$

Then $L_{1}$ must satisfy

$$a x ^ {2} + b m ^ {2} + c n ^ {2} + 2 f m n + 2 g n l + 2 h l m = 0$$

Now the plane $\Pi \to \langle p - p_0,\vec{v}\rangle$ with

$$p _ {0} = (0, 0, 0)$$ $$p = (x, y, z)$$ $$\vec {v} = (l, m, n)$$

is orthogonal to $L_{1}$

This plane cuts $C$ in two other lines $(L_2, L_3)$ such that $L_2 \perp L_3$ if

$$(a + b + c) (l ^ {2} + m ^ {2} + n ^ {2}) - C (l, m, n) = 0$$

or

$$(a + b + c) (l ^ {2} + m ^ {2} + n ^ {2}) = 0$$

or

$$a + b + c = 0$$

because $l^2 + m^2 + n^2 \neq 0$

So we have

$$\vec{v} = (l, m, n) = (1, 1, -1)$$ $$f = -5, g = 8, h = 3$$

Then solving

$$\left\{ \begin{array}{l} fyz + gzx + hxy = 0 \\ lx + my + nz = 0 \end{array} \right.$$

we obtain $L_2, L_3$ as follows

$$L_2 = \left\{ \begin{array}{l} x = \frac{z}{3} \\ y = \frac{2}{3}z \end{array} \right.$$ $$L_3 = \left\{ \begin{array}{l} x = 5z \\ y = -4z \end{array} \right.$$

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