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Difference between revisions of "2014 AMC 12B Problems/Problem 17"

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<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 52\qquad\textbf{(E)}\ 80 </math>
 
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 52\qquad\textbf{(E)}\ 80 </math>
  
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==Solution (Algebra Based)==
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==Solution 1 (Algebra Based)==
 
Let <math>y = m(x - 20) + 14</math>. Equating them:
 
Let <math>y = m(x - 20) + 14</math>. Equating them:
  

Revision as of 18:17, 17 April 2020

Problem

Let $P$ be the parabola with equation $y=x^2$ and let $Q = (20, 14)$. There are real numbers $r$ and $s$ such that the line through $Q$ with slope $m$ does not intersect $P$ if and only if $r$ < $m$ < $s$. What is $r + s$?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 52\qquad\textbf{(E)}\ 80$

Solution 1 (Algebra Based)

Let $y = m(x - 20) + 14$. Equating them:

$x^2 = mx - 20m + 14$

$x^2 - mx + 20m - 14 = 0$

For there to be no solutions, the discriminant must be less than zero:

$m^2 - 4(20m - 14) < 0$

$m^2 - 80m + 56 < 0$.

So $m < 0$ for $r < m < s$ where $r$ and $s$ are the roots of $m^2 - 80m + 56 = 0$ and their sum by Vieta's formulas is $\boxed{\textbf{(E)}\ 80}$.

Solution 2 (Calculus-based)

The line will begin to intercept the parabola when its slope equals that of the parabola at the point of tangency. Taking the derivative of the equation of the parabola, we get that the slope equals $2x$. Using the slope formula, we find that the slope of the tangent line to the parabola also equals $\frac{14-x^2}{20-x}$. Setting these two equal to each other, we get \[2x = \frac{14-x^2}{20-x} \implies x^2-40x+14 = 0\] Solving for $x$, we get \[x= 20\pm \sqrt{386}\] The sum of the two possible values for $x$ where the line is tangent to the parabola is $40$, and the sum of the slopes of these two tangent lines is equal to $2x$, or $\boxed{\textbf{(E)}\ 80}$.

See also

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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 12 Problems and Solutions

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