What type of conic section has the equation $4x^2 - y^2 +4y - 20= 0$ ?

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Answer 1 · 2014-09-25T01:39:52

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

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

$4(x-0)^2-y^2+4y=20$

We need to complete the square using the $y$ term.

$(4/2)^2=(2)^2=4$ , Add $4$ to both sides

$4(x-0)^2-y^2+4y+4=20+4$

$y^2+4y+4 =>(y+2)^2=>$ Perfect square trinomial

Re-write:

$4(x-0)^2-(y+2)^2=20+4$

$4(x-0)^2-(y+2)^2=24$

$(4(x-0)^2)/24-((y+2)^2)/24=24/24$

$((x-0)^2)/6-((y+2)^2)/24=1=>$ Hyperbola

What type of conic section has the equation 4x^2 - y^2 +4y - 20= 04x^2 - y^2 +4y - 20= 0?