Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 18"
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==Problem== | ==Problem== | ||
− | {{ | + | <div style="float:right"> |
+ | [[Image:2006 CyMO-18.PNG|250px]] | ||
+ | </div> | ||
+ | |||
+ | <math>K(k,0)</math> is the minimum point of the parabola and the parabola intersects the y-axis at the point <math>C(0,k)</math>. | ||
+ | If the area if the rectangle <math>OABC</math> is <math>8</math>, then the equation of the parabola is | ||
+ | |||
+ | A. <math>y=\frac{1}{2}(x+2)^2</math> | ||
+ | |||
+ | B. <math>y=\frac{1}{2}(x-2)^2</math> | ||
+ | |||
+ | C. <math>y=x^2+2</math> | ||
+ | |||
+ | D. <math>y=x^2-2x+1</math> | ||
+ | |||
+ | E. <math>y=x^2-4x+4</math> | ||
==Solution== | ==Solution== | ||
− | {{ | + | Since the parabola is symmetric about the line <math>x = k</math>, <math>B</math> has coordinates <math>(2k,k)</math>. The area of the rectangle is <math>k \cdot 2k = 8 \Longrightarrow k = 2</math>, so the vertex is at <math>(2,0)</math>. Thus the equation of the parabola is <math>y = a(x-2)^2</math>, and plugging in point <math>(0,2)</math> and finding <math>a = \frac{1}{2}</math>, the answer is <math>\mathrm{B}</math>. |
==See also== | ==See also== | ||
{{CYMO box|year=2006|l=Lyceum|num-b=17|num-a=19}} | {{CYMO box|year=2006|l=Lyceum|num-b=17|num-a=19}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] |
Revision as of 17:48, 15 October 2007
Problem
is the minimum point of the parabola and the parabola intersects the y-axis at the point
.
If the area if the rectangle
is
, then the equation of the parabola is
A.
B.
C.
D.
E.
Solution
Since the parabola is symmetric about the line ,
has coordinates
. The area of the rectangle is
, so the vertex is at
. Thus the equation of the parabola is
, and plugging in point
and finding
, the answer is
.
See also
2006 Cyprus MO, Lyceum (Problems) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 |