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

Revision as of 21:00, 17 October 2007

Problem

$K(k,0)$ 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 $\mathrm{B}$ .

@@ Line 4: / Line 4: @@
 </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>.
+<math>K(k,0)</math> is the minimum point of the parabola and the parabola intersects the y-axis at the point <math>\Gamma (0,k)</math>.
-If the area if the rectangle <math>OABC</math> is <math>8</math>, then the equation of the parabola is
+If the area if the rectangle <math>OAB\Gamma</math> is <math>8</math>, then the equation of the parabola is
 A. <math>y=\frac{1}{2}(x+2)^2</math>

2006 Cyprus MO, Lyceum (Problems)
Preceded by Problem 17	Followed by Problem 19
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Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 18"

Revision as of 21:00, 17 October 2007

Problem

Solution

See also