Why does the equation 4x225y224x50y+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 hyperbola2(x3)2132(y+1)226=1

3 Answers
May 17, 2015

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: enter image source here

May 17, 2015

Actually, this is not what I have:

4(x26x+99)25(y2+2y+11)+11=0

4(x3)23625(y+1)2+25+11=0

I have that

25+1136=0

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

4(x3)225(y3)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 :)

May 17, 2015

The graph of the equation 4x225y224x50y+11 takes the form of a pair of intersecting lines because the polynomial can be factored as follows:

4x225y224x50y+11 = (2x5y11)(2x+5y1)

enter image source here