Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 18"

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==Problem==
 
==Problem==
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[[Image:2006 CyMO-18.PNG|250px]]
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<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>.
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If the area if the rectangle <math>OABC</math> is <math>8</math>, then the equation of the parabola is
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A. <math>y=\frac{1}{2}(x+2)^2</math>
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B. <math>y=\frac{1}{2}(x-2)^2</math>
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C. <math>y=x^2+2</math>
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D. <math>y=x^2-2x+1</math>
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E. <math>y=x^2-4x+4</math>
  
 
==Solution==
 
==Solution==
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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}}
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[[Category:Introductory Algebra Problems]]

Revision as of 17:48, 15 October 2007

Problem

2006 CyMO-18.PNG

$K(k,0)$ is the minimum point of the parabola and the parabola intersects the y-axis at the point $C(0,k)$. If the area if the rectangle $OABC$ is $8$, then the equation of the parabola is

A. $y=\frac{1}{2}(x+2)^2$

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

C. $y=x^2+2$

D. $y=x^2-2x+1$

E. $y=x^2-4x+4$

Solution

Since the parabola is symmetric about the line $x = k$, $B$ has coordinates $(2k,k)$. The area of the rectangle is $k \cdot 2k = 8 \Longrightarrow k = 2$, so the vertex is at $(2,0)$. Thus the equation of the parabola is $y = a(x-2)^2$, and plugging in point $(0,2)$ and finding $a = \frac{1}{2}$, the answer is $\mathrm{B}$.

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 17
Followed by
Problem 19
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