Difference between revisions of "2014 AMC 12B Problems/Problem 11"
(→Solution) |
(→Solution) |
||
| Line 6: | Line 6: | ||
== Solution == | == Solution == | ||
| − | The line through <math>Q</math> only meets the parabola if the system of equations <math>y = x^2</math> and | + | The line through <math>Q</math> only meets the parabola if the system of equations <math>y = x^2</math> and <math>y = m(x - 20) + 14</math> has one or more solution(s) (the second equation comes from point-slope form). We can equate <math>y</math> and get the equation <math>x^2 = mx - 20m + 14</math>. Now if we move all the terms to the left, we get <math>x^2 - mx - (20m - 14) = 0</math>. We want this equation to have no solutions, and are trying to find the bounds of <math>r</math> and <math>s</math> for <math>m</math> such that this is unsolvable. A quadratic is unsolvable across the reals if its discriminant is less than 0. So, we set the discriminant to be less than 0: |
| + | <cmath>\sqrt{m^2 - 80m - 14} \le 0]</cmath> | ||
Revision as of 19:15, 20 February 2014
Problem
Let 
be the parabola with equation 
and let 
. There are real numbers 
and 
such that the line through 
with slope 
does not intersect if and only if 
. What is 
?
\[\textbf{(A)}\ 1\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}}\ 52\qquad\textbf{(E)}\ 80\] (Error compiling LaTeX. Unknown error_msg)
Solution
The line through only meets the parabola if the system of equations and 
has one or more solution(s) (the second equation comes from point-slope form). We can equate 
and get the equation 
. Now if we move all the terms to the left, we get 
. We want this equation to have no solutions, and are trying to find the bounds of and for such that this is unsolvable. A quadratic is unsolvable across the reals if its discriminant is less than 0. So, we set the discriminant to be less than 0:
![\[\sqrt{m^2 - 80m - 14} \le 0]\]](http://latex.artofproblemsolving.com/a/0/8/a085423a59eb70961c42a3815da2d1a8694377c2.png)
