Question

At what point the origin must be shifted so that linear terms in the conicoid x^2 + 2y^2 - z^2 - 2yz + 2xz + x - 3y + z + 4 = 0 vanish? Justify.

Answer 1

Answer on Question #45097 – Math - Analytic Geometry

Problem.

At what point the origin must be shifted so that linear terms in the conicoid $x^2 + 2y^2 - z^2 - 2yz + 2xz + x - 3y + z + 4 = 0$ vanish? Justify.

Solution.

Suppose that the origin is shifted to point $(a, b, c)$, $(x, y, z)$ are coordinates in old system, $(x', y', z')$ are coordinates in new system. Then $x = x' + a$, $y = y' + b$, $z = z' + c$. After

substation $x = x' + a$, $y = y' + b$, $z = z' + c$ into $x^2 + 2y^2 - z^2 - 2yz + 2xz + x - 3y + z + 4 = 0$ we will obtain

$$\begin{array}{l} (x')^2 + 2x'a + a^2 + 2(y')^2 + 4y'b + 2b^2 - (z')^2 - 2z'c - c^2 - 2y'z' - 2y'c - 2z'b - 2bc \\ + 2x'z' + 2x'c + 2z'a + 2ac + x' + a - 3y' - 3b + z' + c + 4 = 0 \end{array}$$

or

$$\begin{array}{l} (x')^2 + 2(y')^2 - (z')^2 - 2y'z' + 2x'z' + x'(2a + 2c + 1) + y'(4b - 2c - 3) - z'(2c + 2b \\ - 2a - 1) + a^2 + 2b^2 - c^2 - 2bc + 2ac + a - 3b + c + 4 = 0. \end{array}$$

The linear terms should vanish, so

$$\left\{ \begin{array}{l} 2a + 2c + 1 = 0; \\ 4b - 2c - 3 = 0; \\ 2c + 2b - 2a - 1 = 0; \end{array} \right.$$ $$\left\{ \begin{array}{l} 2a + 2c + 1 = 0; \\ 4b - 2c - 3 = 0; \end{array} \right.$$ $$(2c + 2b - 2a - 1) + (2a + 2c + 1) = 2b + 4c = 0.$$ $$\left\{ \begin{array}{l} 2a + 2c + 1 = 0; \\ 4b - 2c - 3 = 0; \end{array} \right.$$ $$b = -2c.$$ $$\left\{ \begin{array}{l} a = -0.2; \\ c = -0.3; \\ b = 0.6. \end{array} \right.$$

Answer: $(-0.2, -0.3, 0.6)$

www.AssignmentExpert.com