Difference between revisions of "1985 AHSME Problems/Problem 19"

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Problem

Consider the graphs $y=Ax^2$ and $y^2+3=x^2+4y$ , where $A$ is a positive constant and $x$ and $y$ are real variables. In how many points do the two graphs intersect?

$\mathrm{(A) \ }\text{exactly }4 \qquad \mathrm{(B) \ }\text{exactly }2 \qquad$

$\mathrm{(C) \ }\text{at least }1,\text{ but the number varies for different positive values of }A \qquad$

$\mathrm{(D) \ }0\text{ for at least one positive value of }A \qquad \mathrm{(E) \ }\text{none of these}$

Solution

Solution 1: Algebra

Substitute $y=Ax^2$ into $y^2+3=x^2+4y$ to get $A^2x^4+3=x^2+4Ax^2\implies A^2x^4-(4A+1)x^2+3=0$ . Let $y=x^2$ , so that $A^2y-(4A+1)y+3=0$ . By the quadratic formula, we have $y=x^2=\frac{4A+1\pm\sqrt{4A^2+8A+1}}{2A^2}$ . Notice that since $A$ is positive, $4A^2+8A+1$ is also positive, so the square root is real. Also, $(4A+1)^2=16A^2+8A+1>4A^2+8A+1$ , so $4A+1-\sqrt{4A^2+8A+1}>0$ , and both solutions to $x^2$ are positive. Thus, we have two possible values for $x^2$ , and taking both the positive and negative roots, we have $4$ possible values for $x$ , $\boxed{\text{A}}$ .

Solution 2: Graphing

Notice that $y=Ax^2$ is an upward-facing parabola with a vertex at the origin. We can manipulate $y^2+3=x^2+4y$ into a recognizable graph:

$y^2+3=x^2+4y$

$y^2-4y+3-x^2=0$

$y^2-4y+4-x^2=1$

$\frac{(y-2)^2}{1}-\frac{(x-0)^2}{1}=1$

This is just a vertical hyperbola! The center of the hyperbola is $(0, 2)$ and it has asymptotes at $y=x+1$ and $y=-x+1$ . Now we can see where a parabola would intersect this hyperbola. It would seem obvious that the parabola intersects the lower branch of the hyperbola twice. For a more rigorous argument, consider that since the y value is always positive, the entire parabola would have to be contained in that area between the lower branch and the x-axis, clearly a violation of the end behavior of approaching infinity.

As for the top branch, it may seem to make sense that a small enough $A$ would make a 'flat' enough parabola so that it doesn't intersect the top branch. However, this is not the case. Consider that as the branch approaches its asymptote, its slope also approaches that of its asymptote, which is $1$ . However, for any parabola, the slope approaches infinity as $x$ does, so eventually any upward-facing parabola will overtake the hyperbola on both sides, for $2$ intersection points on the top branch, too.

Thus, we have a total of $2+2=4$ intersection points, $\boxed{\text{A}}$ .

@@ Line 25: / Line 25: @@
 This is just a vertical hyperbola! The center of the hyperbola is <math> (0, 2) </math> and it has asymptotes at <math> y=x+1 </math> and <math> y=-x+1 </math>. Now we can see where a parabola would intersect this hyperbola. It would seem obvious that the parabola intersects the lower branch of the hyperbola twice. For a more rigorous argument, consider that since the y value is always positive, the entire parabola would have to be contained in that area between the lower branch and the x-axis, clearly a violation of the end behavior of approaching infinity.
-As for the top branch, it may seem to make sense that a small enough <math> A </math> would make a 'flat' enough paraboloa so that it doesn't intersect the top branch. However, this is not the case. Consider that as the branch approaches its asymptote, its slope also approaches that of its asymptote, which is <math> 1 </math>. However, for any parabola, the slope approaches infinity as <math> x </math> does, so eventually any upward-facing parabola will overtake the hyperbola on both sides, for <math> 2 </math> intersection points on the top branch, too.
+As for the top branch, it may seem to make sense that a small enough <math> A </math> would make a 'flat' enough parabola so that it doesn't intersect the top branch. However, this is not the case. Consider that as the branch approaches its asymptote, its slope also approaches that of its asymptote, which is <math> 1 </math>. However, for any parabola, the slope approaches infinity as <math> x </math> does, so eventually any upward-facing parabola will overtake the hyperbola on both sides, for <math> 2 </math> intersection points on the top branch, too.
 Thus, we have a total of <math> 2+2=4 </math> intersection points, <math> \boxed{\text{A}} </math>.

1985 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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