a.
To prove: The graph of the equation asa circle.
a.
Answer to Problem 7PS
It is a circle
Explanation of Solution
Given:
Conic | Condition | |
a | Circle | A=C,A≠0 |
b | Parabola | A=0or C=0, but not both. |
c | Ellipse | AC>0,A≠C |
d | Hyperbola | AC<0 |
Calculation:
By completing square method,
The standard form of equation of a cirlce is.
Hence, proved that the given equation as a circle.
b.
To prove: The graph of the equation as a parabola.
b.
Answer to Problem 7PS
For parabola
Explanation of Solution
Given:
Conic | Condition | |
a | Circle | A=C,A≠0 |
b | Parabola | A=0or C=0, but not both. |
c | Ellipse | AC>0,A≠C |
d | Hyperbola | AC<0 |
By completing square method,
The standard form of equation of a cirlce is.
Hence, proved that the given equation as a parabola.
c.
To prove: The graph of the equation as an ellipse.
c.
Answer to Problem 7PS
For the circle the given equation is
Explanation of Solution
Given:
Conic | Condition | |
a | Circle | A=C,A≠0 |
b | Parabola | A=0or C=0, but not both. |
c | Ellipse | AC>0,A≠C |
d | Hyperbola | AC<0 |
By completing square method,
The standard form of equation of anellipse is.
Hence, proved that the given equation as anellipse.
d.
To prove: The graph of the equation as a hyperbola.
d.
Answer to Problem 7PS
For the circle the given equation is
Explanation of Solution
Given:
Conic | Condition | |
a | Circle | A=C,A≠0 |
b | Parabola | A=0or C=0, but not both. |
c | Ellipse | AC>0,A≠C |
d | Hyperbola | AC<0 |
Calculation:
By completing square method,
The standard form of equation of a hyperbola is.
Hence, proved that the given equation as a hyperbola.
Chapter 10 Solutions
EBK PRECALCULUS W/LIMITS
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