# Why does the equation 4x^2-25y^2-24x-50y+11=0 not take the form of a hyperbola, despite the fact that the squared terms of the equation have different signs? Also, why can this equation be put in the form of hyperbola(2(x-3)^2)/13 - (2(y+1)^2)/26=1

May 17, 2015

Also, here is the work for getting the equation into the form of a hyperbola:

May 17, 2015

Actually, this is not what I have:

$4 \left({x}^{2} - 6 x + 9 - 9\right) - 25 \left({y}^{2} + 2 y + 1 - 1\right) + 11 = 0 \implies$

$\implies 4 {\left(x - 3\right)}^{2} - 36 - 25 {\left(y + 1\right)}^{2} + 25 + 11 = 0$

I have that

$25 + 11 - 36 = 0$

so it's a reducible conic whose polynomial has real roots

$4 {\left(x - 3\right)}^{2} - 25 {\left(y - 3\right)}^{2} = 0$

So it splits up in 2 real-valued lines which intersecate in the center $\left(3 , - 1\right)$

The first statement is only necessary to have an hyperbola: you need also the equation not to be reducible, or you have a degenerate conic.

Check your calculations, and don't worry, everybody makes mistakes in calculations :)

May 17, 2015

The graph of the equation $4 {x}^{2} - 25 {y}^{2} - 24 x - 50 y + 11$ takes the form of a pair of intersecting lines because the polynomial can be factored as follows:

$4 {x}^{2} - 25 {y}^{2} - 24 x - 50 y + 11$ $=$ $\left(2 x - 5 y - 11\right) \left(2 x + 5 y - 1\right)$