A determinant and a locus

Algebra Level pending

Classify the curve r = ( x , y ) \mathbf{r} = (x, y) , where x x and y y satisfy the determinant equation

∣ x 1 2 0 2 1 1 y 1 ∣ = 0 \begin{vmatrix} x &&1 &&2 \\ 0 && 2 && 1 \\ 1 && y && 1 \end{vmatrix} = 0

Straight Line A Single Point Hyperbola Circle Parabola

Hosam Hajjir by Hosam Hajjir , 56, Canada

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1 solution

David Vreken
Nov 18, 2020

The determinant calculates to 2 x + 1 − 4 − x y = 0 2x + 1 - 4 - xy = 0 , which after rearranging is x y − 2 x + 3 = 0 xy - 2x + 3 = 0 .

Using the equations found here , A = 0 A = 0 , B = 1 B = 1 , C = 0 C = 0 , a = 0 a = 0 , b = 0 b = 0 , c = 3 c = 3 , f = 0 f = 0 , g = − 1 g = -1 , and h = 1 2 h = \frac{1}{2} .

Therefore, a b c + 2 f g h − a f 2 − b g 2 − c h 2 = − 3 4 ≠ 0 abc + 2fgh - af^2 - bg^2 - ch^2 = -\frac{3}{4} \neq 0 , so the conic is non-degenerate, and B 2 − 4 A C = 1 > 0 B^2 - 4AC = 1 > 0 , so the conic is a hyperbola .

3 Helpful 1 Interesting 1 Brilliant 0 Confused

Not just y = 2 − 3 x y=2-\frac{3}{x} ?

Chris Lewis - 6 months, 3 weeks ago

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That, too. I guess it depends on if you want to put the work into rearranging the equation or into plugging the values in some set equations.

David Vreken - 6 months, 3 weeks ago

This transformation makes it obvious it's a hyperbola, certainly.

Richard Desper - 6 months, 3 weeks ago

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