Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 18"
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==Problem== | ==Problem== | ||
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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>\Gamma (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>OAB\Gamma</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 | ||
− | + | <math>\mathrm{(A)}\ y=\frac{1}{2}(x+2)^2\qquad\mathrm{(B)}\ y=\frac{1}{2}(x-2)^2\qquad\mathrm{(C)}\ y=x^2+2\qquad\mathrm{(D)}\ y=x^2-2x+1\qquad\mathrm{(E)}\ y=x^2-4x+4</math> | |
− | B. <math> | + | ==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>. Plugging in point <math>(0,2)</math>, we find <math>a = \frac{1}{2}</math>, and the answer is <math>\mathrm{B}</math>. | |
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==See also== | ==See also== |
Latest revision as of 12:39, 26 April 2008
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
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 . Plugging in point , we find , and the answer is .
See also
2006 Cyprus MO, Lyceum (Problems) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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