Question

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`

Answer 1

To people, answering the question, please note this graph: https://www.desmos.com/calculator/jixsqaffyw

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

Answer 2

Actually, this is not what I have:

`4(x^2-6x +9 - 9) -25(y^2+2y +1 -1) +11 = 0 =>`

`=> 4(x-3)^2-36-25(y+1)^2+25+11=0`

I have that

`25+11-36=0`

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

`4(x-3)^2-25(y-3)^2=0`

So it splits up in 2 real-valued lines which intersecate in the center `(3,-1)`

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 :)

Answer 3

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` `=` `(2 x - 5 y - 11)(2 x + 5 y - 1)`