McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
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Exercise 57 Page 649

Use the discriminant of the equation by rewriting it in the form Ax^2+Bxy+Cy^2+Dx+Ey+F=0.

Hyperbola

Practice makes perfect

We want to state whether the graph of the given equation is a parabola, a circle, an ellipse, or a hyperbola. To do so, we will rewrite it in the form Ax^2+ Bxy+ Cy^2+Dx+Ey+F=0. Then we can determine its discriminant. Recall that missing terms have coefficient 0. 5y^2+2y+4x-13x^2=81 ⇕ - 13x^2 + 0xy+ 5y^2 + 4x + 2y + (- 81)=0The discriminant of the equation can be found by evaluating B^2-4 A C. This will tell us which type of conic section we might have.

Discriminant Conic Section
B^2-4AC<0; B=0 and A=C Circle
B^2-4AC<0; either B≠ 0 or A≠ C Ellipse
B^2-4AC=0 Parabola
B^2-4AC>0 Hyperbola

Let's calculate the discriminant of our equation.

B^2-4AC
0^2-4( - 13)( 5)
â–¼
Simplify
0-4(- 13)(5)
0+4(13)(5)
0+260
260

The discriminant is greater than 0, so the graph of the equation is a hyperbola.