Calculus- Early Transcendentals, 2021a

28 Review
Determining the Type of Conic
An equation of the form
Ax 2 + Bxy +Cy 2 + Dx + Ey+ F = 0
gives rise to a graph that can be generated by performing a conic section (parabolas, circles, ellipses,
hyperbolas). Note that the Bxy term involves conic rotation. The Dx, Ex,andF terms affect the
vertex and centre. For simplicity, we will omit the Bxy term. To determine the type of graph we
focus our analysis on the values of A and C.
•IfA = C, the graph is a circle.
•IfAC > 0(andA ≠ C), the graph is an ellipse.
•IfAC = 0, the graph is a parabola.
•IfAC < 0, the graph is a hyperbola.
Example 1.29: Center and Radius of a Circle
Find the centre and radius of the circle y 2 + x 2 − 12x + 8y + 43 = 0.
Solution. We need to complete the square twice, once for the x terms and once for the y terms. We’ll do
both at the same time. First let’s collect the terms with x together, the terms with y together, and move the
number to the other side.
(x 2 − 12x)+(y 2 + 8y)=−43
Weadd36tobothsidesforthex term (−12 → −12
2
= −6 → (−6) 2 = 36), and 16 to both sides for the y
term (8 → 8 2 = 4 → (4)2 = 16):
(x 2 − 12x + 36)+(y 2 + 8y + 16)=−43 + 36 + 16
Factoring gives:
(x − 6) 2 +(y + 4) 2 = 3 2 .
Therefore, the centre of the circle is (6,−4) and the radius is 3.
♣
Example 1.30: Type of Conic
What type of conic is 4x 2 − y 2 − 8x + 8 = 0? Put it in standard form.
Solution. Here we have A = 4andC = −1. Since AC < 0, the conic is a hyperbola. Let us complete the
square for the x and y terms. First let’s collect the terms with x together, the terms with y together, and
move the number to the other side.
(4x 2 − 8x) − y 2 = −8
Now we factor out 4 from the x terms.
4(x 2 − 2x) − y 2 = −8