Difference between revisions of "1985 AHSME Problems/Problem 19"
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==Problem== | ==Problem== | ||
− | 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? | + | 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> | ||
− | + | ==Solution 1== | |
− | + | 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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− | + | ==Solution 2== | |
− | + | 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. | |
− | <math> | + | 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. |
− | <math> y | + | 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. |
− | + | Thus there are a total of <math>2+2 = \boxed{\text{(A) exactly} \ 4}</math> intersection points. | |
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− | Thus | ||
==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
Contents
[hide]Problem
Consider the graphs of and , where is a positive constant and and are real variables. In how many points do the two graphs intersect?
Solution 1
Substituting into the equation gives Now observe that since is positive, is also positive, so the square root will always give two distinct real values. Moreover, so , meaning that both solutions for are positive. Hence both solutions will give distinct values of (the positive and negative square roots), and each of these will correspond to a distinct point of intersection of the graphs, so there are points of intersection.
Solution 2
Firstly, note that is an upward-facing parabola (since ) whose vertex is at the origin. We now manipulate the equation of the second graph as follows: showing that it is a vertical (upward- and downward-opening) hyperbola with center and asymptotes and . It therefore remains to consider graphically where the parabola will intersect the hyperbola.
On the lower branch of the hyperbola, the maximum point is , which is above the vertex of the parabola. Therefore, by continuity and the symmetry of both the parabola and the hyperbola in the -axis, there are always exactly intersection points here.
For the top branch, as it approaches the asymptote , its slope also approaches that of this asymptote, which is . However, for any upward-opening parabola, the slope approaches infinity as does, so no matter how small is (i.e. how 'flat' the parabola is), the parabola will eventually overtake the hyperbola, giving a point of intersection with positive -coordinate. As above, symmetry gives another point of intersection with negative -coordinate, so that there are intersection points with this branch too.
Thus there are a total of intersection points.
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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All AHSME Problems and Solutions |
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