Question

What type of conic section has the equation 4x2−y2+4y−20=0?

Answer 1

4x^2-y^2+4y-20=0

4x^2=y^2-4y+20

4x^2=(y-2)^2+15

4x^2-(y-2)^2=15

4x^2/15-(y-2)^2/15=1

therefore its a hyperbola since its of the form x^2/a^2-y^2/b^2=1

where a^2=b^2=15