
Concept explainers
a.
To find:the value of C for no points of intersection.
a.

Answer to Problem 77E
The value of C for no points of intersection is C<−174 and C>2 .
Explanation of Solution
Given information:
The given equation of circle is x2+y2=4 . And the circle and parabola can have 0,1,2,3,4 points of intersection.
The graph for the circle x2+y2=4 is shown in figure (1).
Figure (1)
Calculation:
For no points of intersection between the circle and the parabola.
Substitute the value of x2 from equation of parabola into the equation of circle we get,
y2+y−C−4=0
Th above
(1)2−4×1×(−(C+4))<04C+17<0C<−174......(1)
Substitute the value of y from equation of parabola into circle.
x2+(x2+C)2=4x4+(2C+1)x2+C2−4=0
Above biquadratic equation Should have no real roots. So, its discriminant should be negative.
(2C+1)2−4×1×(C2−4)<04C+8<0C<−2......(2)
From equation of the parabola.
x2=y−C
As x2 is always greater than or equal to zero.
y−C≥0y≥0......(3)
If y is always greater than 2 , then there will be no point of intersection.
From equation (3).
C>2
Then for no point of intersection C<−174 and C>2 .
The graph is shown in below figure (2).
Figure (2)
Therefore, the value of C for no points of intersection is C<−174 and C>2 .
b.
To find:the value of C for one points of intersection.
b.

Answer to Problem 77E
The value of C for one points of intersection is C=2 .
Explanation of Solution
Given information:
The given equation of circle is x2+y2=4 . And the circle and parabola can have 0,1,2,3,4 points of intersection.
The graph for the circle x2+y2=4 is shown in figure (1).
Figure (1)
Calculation:
For one points of intersection the parabola should touch the circle tangentially at one point
This can be possible when the vertex of the parabola touches the circle. And parabola must be tangent to the circle at y=2 .
From parabola equation.
For y=2 and x=0
C=2
The graph is shown in below figure (3).
Figure (3)
Therefore, the value of C for one points of intersection is C=2 .
c.
To find:the value of C for two points of intersection.
c.

Answer to Problem 77E
The value of C for two points of intersection are −2<C<2 or C=−174 .
Explanation of Solution
Given information:
The given equation of circle is x2+y2=4 . And the circle and parabola can have 0,1,2,3,4 points of intersection.
Figure (1)
Calculation:
When we decrease the value of C from C=2 , we will get two point of intersection.
From equation (1) and (2) we get two threshold values of C
−2 and −174 .
When we decrease the value of C from C=2 , a point will reach when we will get three solutions and this point occurs when C=−2 . Also when we cross this threshold we will get four solutions and as soon as C reaches −174 , we will get two point of intersection.
Two point of intersection are −2<C<2 or C=−174 .
The graph is shown in below figure (4).
Figure (4)
Therefore, the value of C for two points of intersection are −2<C<2 or C=−174 .
d.
To find:the value of C for three points of intersection.
d.

Answer to Problem 77E
The value of C for three points of intersection are C=−2 .
Explanation of Solution
Given information:
The given equation of circle is x2+y2=4 . And the circle and parabola can have 0,1,2,3,4 points of intersection.
Calculation:
When we decrease the value of C from C=2 , we will get two point of intersection.
From equation (1) and (2) we get two threshold values of C
−2 and −174 .
When we decrease the value of C from C=2 , a point will reach when we will get three solutions and this point occurs when C=−2 . Also when we cross this threshold we will get four solutions and as soon as C reaches −174 , we will get two point of intersection.
Three point of intersection is C=−2 .
The graph is shown in below figure (5).
Figure (5)
Therefore, the value of C for three points of intersection are C=−2 .
e.
To find:the value of C for four points of intersection.
e.

Answer to Problem 77E
The value of C for four points of intersection are −2<C<−174 .
Explanation of Solution
Given information:
The given equation of circle is x2+y2=4 . And the circle and parabola can have 0,1,2,3,4 points of intersection.
Calculation:
When we decrease the value of C from C=2 , we will get two point of intersection.
From equation (1) and (2) we get two threshold values of C
−2 and −174 .
When we decrease the value of C from C=2 , a point will reach when we will get three solutions and this point occurs when C=−2 . Also when we cross this threshold we will get four solutions and as soon as C reaches −174 , we will get two point of intersection.
Four point of intersection is −2<C<−174 .
The graph is shown in below figure (6).
Figure (6)
Therefore, the value of C for four points of intersection are −2<C<−174 .
Chapter 10 Solutions
EBK PRECALCULUS W/LIMITS
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