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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

1

This post received KUDOS

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]
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?

greprepclubot

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