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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]  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 [ 10.06 KiB | Viewed 2513 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?

[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]  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.
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Re: The figure shows the graph of the equation y =k–x^2, where k [#permalink]  07 Jan 2019, 05:54
IlCreatore wrote:

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.

Is it OC or AC?
Re: The figure shows the graph of the equation y =k–x^2, where k   [#permalink] 07 Jan 2019, 05:54
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