Difference between revisions of "2006 AMC 10A Problems/Problem 8"
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=== Solution 2 === | === Solution 2 === | ||
− | Alternatively, notice that since the equation is that of a | + | Alternatively, notice that since the equation is that of a conic parabola, the vertex is likely <math>(3,2)</math>. Thus, the form of the equation of the parabola is <math>y - 2 = (x - 3)^2</math>. Expanding this out, we find that <math>c = 11</math>. |
=== Solution 3 === | === Solution 3 === |
Revision as of 20:21, 16 July 2019
Contents
Problem
A parabola with equation passes through the points and . What is ?
Solution
Solution 1
Substitute the points and into the given equation for .
Then we get a system of two equations:
Subtracting the first equation from the second we have:
Then using in the first equation:
is the answer.
Solution 2
Alternatively, notice that since the equation is that of a conic parabola, the vertex is likely . Thus, the form of the equation of the parabola is . Expanding this out, we find that .
Solution 3
The points given have the same -value, so the vertex lies on the line .
The -coordinate of the vertex is also equal to , so set this equal to and solve for , given that :
Now the equation is of the form . Now plug in the point and solve for :
Solution 4
Substituting y into the two equations, we get:
Which can be written as:
4, 2, are the solutions to the quadratic. Thus:
See also
2006 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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