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The figure shows the graph of the equation y =k–x^2, where k

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The figure shows the graph of the equation y =k–x^2, where k [#permalink] New post 29 Oct 2017, 01:27
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0% (00:00) correct 100% (01:59) wrong based on 2 sessions
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Coordinate Geometry - area- NE question.png
Coordinate Geometry - area- NE question.png [ 10.06 KiB | Viewed 1791 times ]

The figure shows the graph of the equation y =k–x^2, where k is a constant. If the area of triangle ABC is 1/8, what is the value of k?

Give your answer to the nearest 0.01

Enter your value.

[Reveal] Spoiler: OA
0.25
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Re: The figure shows the graph of the equation y =k–x^2, where k [#permalink] New post 31 Oct 2017, 00:55
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In order to find the coordinates of the vertices of the triangle, let's exploit the equation of the parabola.

To find the coordinates of points B and C, let's evaluate the parabola at y = 0, i.e. \(0 = -x^2+k\) which leads to \(x=\pm\sqrt{k}\).

Then, vertices A has the same coordinates of the vertex of the parabola, i.e. evaluating at x = 0, we get \(y=k\).

Then, we can impose an equation for the area of the triangle as follows \(\frac{BC*OC}{2} = \frac{2\sqrt{k}*k}{2} =\frac{1}{8}\). Solving for k, we get \(k\sqrt{k}=\sqrt{k^3}=\frac{1}{8}\), i.e. k = 1/4 = 0.25.
Re: The figure shows the graph of the equation y =k–x^2, where k   [#permalink] 31 Oct 2017, 00:55
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The figure shows the graph of the equation y =k–x^2, where k

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