2006 Cyprus MO/Lyceum/Problem 18

Revision as of 22:00, 17 October 2007 by 1=2 (talk | contribs) (Problem)

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 $\Gamma (0,k)$. If the area if the rectangle $OAB\Gamma$ 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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