How do use the discriminant test to determine whether the graph $x^2+5xy+1-y^2-16=0$ whether the graph is parabola, ellipse, or hyperbola?

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Answer 1 · 2018-07-20T06:57:29

It is a hyperbola.

Let the equation be of the type $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$

then if

$B^2-4AC=0$ and $A=0$ or $C=0$ , it is a parabola

$B^2-4AC<0$ and $A=C$ , it is a circle

$B^2-4AC<0$ and $A!=C$ , it is an ellipse

$B^2-4AC>0$ , it is a hyperbola

In the given equation $x^2+5xy+1-y^2-16=0$ or $x^2+5xy-y^2-15=0$

$A=1$ , $B=5$ and $C=-1$

Therefore, $B^2-4AC=25+4=29$

Hence, it is a hyperbola.

graph{x^2+5xy+1-y^2-16=0 [-20, 20, -10, 10]}

How do use the discriminant test to determine whether the graph x^2+5xy+1-y^2-16=0x^2+5xy+1-y^2-16=0 whether the graph is parabola, ellipse, or hyperbola?