Question

For what value(s) of α, the conicoid x^2+y^2+az^2+2yz +xy+2y+z+3=0 has a unique centre? Give reason for your answer.

Answer 1

Answer on Question #46165 – Math – Analytic Geometry

Problem.

For what value(s) of $\alpha$, the conicoid $x^2 + y^2 + \alpha z^2 + 2yz + xy + 2y + z + 3 = 0$ has a unique centre? Give reason for your answer.

Solution.

Denoting the given expression by $F(x,y,z)$ we get from

$$\frac{\partial F}{\partial x} = 0, \frac{\partial F}{\partial y} = 0, \frac{\partial F}{\partial z} = 0$$

or

$$2x + y = 0,$$ $$2y + 2z + x + 2 = 0,$$ $$2\alpha z + 2y + 1 = 0.$$

Let $A = \begin{bmatrix} 2 & 1 & 0 \\ 1 & 2 & 2 \\ 0 & 2 & 2\alpha \end{bmatrix}$ and $A' = \begin{bmatrix} 2 & 1 & 0 & 0 \\ 1 & 2 & 2 & 2 \\ 0 & 2 & 2\alpha & 1 \end{bmatrix}$.

The conicoid $x^2 + y^2 + \alpha z^2 + 2yz + xy + 2y + z + 3 = 0$ has a unique centre if $\text{rank } A = \text{rank } A' = 3$.

$$A = \begin{bmatrix} 2 & 1 & 0 \\ 1 & 2 & 2 \\ 0 & 2 & 2\alpha \end{bmatrix} \sim \begin{bmatrix} 0 & -3 & -4 \\ 1 & 2 & 2 \\ 0 & 2 & 2\alpha \end{bmatrix} \sim \begin{bmatrix} 1 & 2 & 2 \\ 0 & -3 & -4 \\ 0 & 2 & 2\alpha \end{bmatrix} \sim \begin{bmatrix} 1 & 2 & 2 \\ 0 & -3 & -4 \\ 0 & 0 & 6\alpha - 8 \end{bmatrix}$$

Hence $\text{rank } A = 3$ if $\alpha \neq \frac{4}{3}$.

$$A' = \begin{bmatrix} 2 & 1 & 0 & 0 \\ 1 & 2 & 2 & 2 \\ 0 & 2 & 2\alpha & 1 \end{bmatrix} \sim \begin{bmatrix} 0 & -3 & -4 & -4 \\ 1 & 2 & 2 & 2 \\ 0 & 2 & 2\alpha & 1 \end{bmatrix} \sim \begin{bmatrix} 1 & 2 & 2 & 2 \\ 0 & -3 & -4 & -4 \\ 0 & 2 & 2\alpha & 1 \end{bmatrix} \sim \begin{bmatrix} 1 & 2 & 2 & 2 \\ 0 & -3 & -4 & -4 \\ 0 & 0 & 6\alpha - 8 & -5 \end{bmatrix}$$

Hence $\text{rank } A' = 3$.

Therefore conicoid has a unique centre if $\alpha \neq \frac{4}{3}$.

Answer: $\alpha \neq \frac{4}{3}$.

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