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Difference between revisions of "1985 AHSME Problems/Problem 19"
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==Problem==
 
==Problem==
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Consider the graphs <math> y=Ax^2 </math> and <math> y^2+3=x^2+4y </math>, where <math> A </math> is a positive constant and <math> x </math> and <math> y </math> are real variables. In how many points do the two graphs intersect?
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Consider the graphs of <math>y = Ax^2</math> and <math>y^2+3 = x^2+4y</math>, where <math>A</math> is a positive constant and <math>x</math> and <math>y</math> are real variables. In how many points do the two graphs intersect?
  
 
<math> \mathrm{(A) \ }\text{exactly }4 \qquad \mathrm{(B) \ }\text{exactly }2 \qquad </math>  
 
<math> \mathrm{(A) \ }\text{exactly }4 \qquad \mathrm{(B) \ }\text{exactly }2 \qquad </math>  
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<math>  \mathrm{(D) \  }0\text{ for at least one positive value of }A \qquad \mathrm{(E) \ }\text{none of these} </math>
 
<math>  \mathrm{(D) \  }0\text{ for at least one positive value of }A \qquad \mathrm{(E) \ }\text{none of these} </math>
  
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==Solution==
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==Solution 1==
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===Solution 1: Algebra===
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Substituting <math>y = Ax^2</math> into the equation <math>y^2+3 = x^2+4y</math> gives <cmath>\begin{align*}\left(Ax^2\right)+3 = x^2+4\left(Ax^2\right) &\iff A^2x^4+3 = x^2+4Ax^2 \\ &\iff A^2x^4-\left(4A+1\right)x^2+3 = 0 \\ &\iff x^2 = \frac{4A+1 \pm \sqrt{4A^2+8A+1}}{2A^2} \\ &\text{(using the quadratic formula)}.\end{align*}</cmath> Now observe that since <math>A</math> is positive, <math>4A^2+8A+1</math> is also positive, so the square root will always give two distinct real values. Moreover, <cmath>\left(4A+1\right)^2 = 16A^2+8A+1 > 4A^2+8A+1,</cmath> so <math>4A+1-\sqrt{4A^2+8A+1} > 0</math>, meaning that both solutions for <math>x^2</math> are positive. Hence both solutions will give <math>2</math> distinct values of <math>x</math> (the positive and negative square roots), and each of these will correspond to a distinct point of intersection of the graphs, so there are <math>2 \cdot 2 = \boxed{\text{(A) exactly} \ 4}</math> points of intersection.
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Substitute <math> y=Ax^2 </math> into <math> y^2+3=x^2+4y </math> to get <math> A^2x^4+3=x^2+4Ax^2\implies A^2x^4-(4A+1)x^2+3=0 </math>. Let <math> y=x^2 </math>, so that <math> A^2y-(4A+1)y+3=0 </math>. By the quadratic formula, we have <math> y=x^2=\frac{4A+1\pm\sqrt{4A^2+8A+1}}{2A^2} </math>. Notice that since <math> A </math> is positive, <math> 4A^2+8A+1 </math> is also positive, so this is not complex. Also, <math> (4A+1)^2=16A^2+8A+1>4A^2+8A+1 </math>, so <math> 4A+1-\sqrt{4A^2+8A+1}>0 </math>, and both solutions to <math> x^2 </math> are positive. Thus, we have two possible values for <math> x^2 </math>, and taking both the positive and negative roots, we have <math> 4 </math> possible values for <math> x </math>, <math> \boxed{\text{A}} </math>.
 
  
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===Solution 2: Graphing===
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==Solution 2==
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Notice that <math> y=Ax^2 </math> is an upward-facing parabola with a vertex at the origin. We can manipulate <math> y^2+3=x^2+4y </math> into a recognizable graph:
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Firstly, note that <math>y = Ax^2</math> is an upward-facing parabola (since <math>A > 0</math>) whose vertex is at the origin. We now manipulate the equation of the second graph as follows: <cmath>\begin{align*}y^2+3 = x^2+4y &\iff y^2-4y+3-x^2 = 0 \\ &\iff y^2-4y+4-x^2 = 1 \\ &\iff \frac{(y-2)^2}{1}-\frac{(x-0)^2}{1} = 1,\end{align*}</cmath> showing that it is a vertical (upward- and downward-opening) hyperbola with center <math>(0,2)</math> and asymptotes <math>y=x+1</math> and <math>y=-x+1</math>. It therefore remains to consider graphically where the parabola will intersect the hyperbola.
  
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<math> y^2+3=x^2+4y </math>
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On the lower branch of the hyperbola, the maximum point is <math>(0,2-1) = (0,1)</math>, which is above the vertex of the parabola. Therefore, by continuity and the symmetry of both the parabola and the hyperbola in the <math>y</math>-axis, there are always exactly <math>2</math> intersection points here.
  
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<math> y^2-4y+3-x^2=0 </math>
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For the top branch, as it approaches the asymptote <math>y = x+1</math>, its slope also approaches that of this asymptote, which is <math>1</math>. However, for any upward-opening parabola, the slope approaches infinity as <math>x</math> does, so no matter how small <math>A</math> is (i.e. how 'flat' the parabola is), the parabola will eventually overtake the hyperbola, giving a point of intersection with positive <math>x</math>-coordinate. As above, symmetry gives another point of intersection with negative <math>x</math>-coordinate, so that there are <math>2</math> intersection points with this branch too.
  
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<math> y^2-4y+4-x^2=1 </math>
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Thus there are a total of <math>2+2 = \boxed{\text{(A) exactly} \ 4}</math> intersection points.
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<math> \frac{(y-2)^2}{1}-\frac{(x-0)^2}{1}=1 </math>
 
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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.
 
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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.
 
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Thus, we have a total of <math> 2+2=4 </math> intersection points, <math> \boxed{\text{A}} </math>.
 
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=18|num-a=20}}
 
{{AHSME box|year=1985|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:10, 19 March 2024

Problem

Consider the graphs of $y = Ax^2$ 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 1

Substituting $y = Ax^2$ into the equation $y^2+3 = x^2+4y$ gives \begin{align*}\left(Ax^2\right)+3 = x^2+4\left(Ax^2\right) &\iff A^2x^4+3 = x^2+4Ax^2 \\ &\iff A^2x^4-\left(4A+1\right)x^2+3 = 0 \\ &\iff x^2 = \frac{4A+1 \pm \sqrt{4A^2+8A+1}}{2A^2} \\ &\text{(using the quadratic formula)}.\end{align*} Now observe that since $A$ is positive, $4A^2+8A+1$ is also positive, so the square root will always give two distinct real values. Moreover, \[\left(4A+1\right)^2 = 16A^2+8A+1 > 4A^2+8A+1,\] so $4A+1-\sqrt{4A^2+8A+1} > 0$, meaning that both solutions for $x^2$ are positive. Hence both solutions will give $2$ distinct values of $x$ (the positive and negative square roots), and each of these will correspond to a distinct point of intersection of the graphs, so there are $2 \cdot 2 = \boxed{\text{(A) exactly} \ 4}$ points of intersection.

Solution 2

Firstly, note that $y = Ax^2$ is an upward-facing parabola (since $A > 0$) whose vertex is at the origin. We now manipulate the equation of the second graph as follows: \begin{align*}y^2+3 = x^2+4y &\iff y^2-4y+3-x^2 = 0 \\ &\iff y^2-4y+4-x^2 = 1 \\ &\iff \frac{(y-2)^2}{1}-\frac{(x-0)^2}{1} = 1,\end{align*} showing that it is a vertical (upward- and downward-opening) hyperbola with center $(0,2)$ and asymptotes $y=x+1$ and $y=-x+1$. It therefore remains to consider graphically where the parabola will intersect the hyperbola.

On the lower branch of the hyperbola, the maximum point is $(0,2-1) = (0,1)$, which is above the vertex of the parabola. Therefore, by continuity and the symmetry of both the parabola and the hyperbola in the $y$-axis, there are always exactly $2$ intersection points here.

For the top branch, as it approaches the asymptote $y = x+1$, its slope also approaches that of this asymptote, which is $1$. However, for any upward-opening parabola, the slope approaches infinity as $x$ does, so no matter how small $A$ is (i.e. how 'flat' the parabola is), the parabola will eventually overtake the hyperbola, giving a point of intersection with positive $x$-coordinate. As above, symmetry gives another point of intersection with negative $x$-coordinate, so that there are $2$ intersection points with this branch too.

Thus there are a total of $2+2 = \boxed{\text{(A) exactly} \ 4}$ intersection points.

See Also

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