Difference between revisions of "1984 AHSME Problems/Problem 22"
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==Problem== | ==Problem== | ||
| − | Let <math> a </math> and <math> c </math> be fixed [[Natural numbers|positive numbers]]. For each [[real number]] <math> t </math> let <math> (x_t, y_t) </math> be the [[vertex]] of the [[parabola]] <math> y=ax^2+ | + | Let <math> a </math> and <math> c </math> be fixed [[Natural numbers|positive numbers]]. For each [[real number]] <math> t </math> let <math> (x_t, y_t) </math> be the [[vertex]] of the [[parabola]] <math> y=ax^2+tx+c </math>. If the set of the vertices <math> (x_t, y_t) </math> for all real numbers of <math> t </math> is graphed on the [[Cartesian plane|plane]], the [[graph]] is |
<math> \mathrm{(A) \ } \text{a straight line} \qquad \mathrm{(B) \ } \text{a parabola} \qquad \mathrm{(C) \ } \text{part, but not all, of a parabola} \qquad \mathrm{(D) \ } \text{one branch of a hyperbola} \qquad </math> <math> \mathrm{(E) \ } \text{None of these} </math> | <math> \mathrm{(A) \ } \text{a straight line} \qquad \mathrm{(B) \ } \text{a parabola} \qquad \mathrm{(C) \ } \text{part, but not all, of a parabola} \qquad \mathrm{(D) \ } \text{one branch of a hyperbola} \qquad </math> <math> \mathrm{(E) \ } \text{None of these} </math> | ||
==Solution== | ==Solution== | ||
| − | The x-coordinate of the vertex of | + | The <math>x</math>-coordinate of the vertex of the parabola <math>ax^2 + tx + c</math> is <math> -\frac{t}{2a} </math>, so <math> x_t=-\frac{t}{2a} </math>. Substituting yields <math>y_t=-\frac{t^2}{4a^2}+c </math>. Notice that <math> y_t=-\frac{t^2}{4a^2}+c=-a(-\frac{t}{2a})^2+c=-ax_t^2+c </math>, so all of the vertices are on a parabola. However, we have only showed that all of the points in the locus of vertices are on a parabola, we have not shown whether or not all points on the parabola are on the locus. Assume we are given an <math> x_t </math> on the parabola. <math> -\frac{t}{2a}=x_t </math>, <math>t = -2ax_t </math>, so a unique <math> t </math>, and therefore a unique vertex, is determined for each point on the parabola. We therefore conclude that every point in the locus is on the parabola and every point on the parabola is in the locus, and the graph of the locus is the same as the graph of the parabola, <math> \boxed{\text{B}} </math>. |
==See Also== | ==See Also== | ||
{{AHSME box|year=1984|num-b=21|num-a=23}} | {{AHSME box|year=1984|num-b=21|num-a=23}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:55, 3 July 2025
Problem
Let 
and 
be fixed positive numbers. For each real number 
let 
be the vertex of the parabola 
. If the set of the vertices for all real numbers of is graphed on the plane, the graph is
Solution
The 
-coordinate of the vertex of the parabola 
is 
, so 
. Substituting yields 
. Notice that 
, so all of the vertices are on a parabola. However, we have only showed that all of the points in the locus of vertices are on a parabola, we have not shown whether or not all points on the parabola are on the locus. Assume we are given an 
on the parabola. 
, 
, so a unique , and therefore a unique vertex, is determined for each point on the parabola. We therefore conclude that every point in the locus is on the parabola and every point on the parabola is in the locus, and the graph of the locus is the same as the graph of the parabola, 
.
See Also
| 1984 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
