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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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Attachment:
Coordinate Geometry  area NE question.png [ 10.06 KiB  Viewed 11623 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 \(\frac{1}{8}\), what is the value of k? Give your answer to the nearest 0.01




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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: GRE Math Challenge #2 (170 level question) [#permalink]
20 Sep 2018, 04:29
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Tough question. The first part of the solution:Let the origin be O. The area of the triangle ABC is 1/8. The area of the triangle ABC = the area of the triangle AOB*2 The area of the triangle AOB = (distance between Origin and B * distance between Origin and A)/2 The area of the triangle ABC = ((distance between Origin and B * distance between Origin and A)/2)*2 => The area of the triangle ABC = distance between Origin and B * distance between Origin and A = 1/8 The second part of the solution:Point A is the point when x=0 If x=0, then y=k0 => y=k So, we now know that the distance between the origin and the point A is k. Points B and C are points when y=0 0=kx^2 => x^2 = k => x = + sqrt(k) So, we now know that the distance between the origin and the point B is sqrt(k)The third part of the solution:k*sqrt(k) = 1/8 k^3 = 1/64 k = 1/sqrt(8) k = 0.35 Kudos if you liked this solution



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The figure shows the graph of the equation y = k – x^2, [#permalink]
11 Dec 2018, 14:11
Attachment:
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The figure shows the graph of the equation \(y = k – x^2\), where k is a constant. If the area of triangle ABC is \(\frac{1}{8}\), what is the value of k? Give your answer to the nearest 0.01
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Re: The figure shows the graph of the equation y = k – x^2, [#permalink]
16 Dec 2018, 04:48
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Carcass wrote: Attachment: #GREpracticequestion The figure shows the graph of the equation.jpg The figure shows the graph of the equation \(y = k – x^2\), where k is a constant. If the area of triangle ABC is \(\frac{1}{8}\), what is the value of k? Give your answer to the nearest 0.01 Explanation::Given \(y = k  x^2\) Now considering the y  axis and x = 0 Therefore, \(y = k  0\) or \(y = k\) So the point A coordinate =\(k,0\) Similarly, for point c, \(y =0\) then,\(y = k x^2\) or\(0 = k  x^2\) or\(x = \sqrt{k}\) therefore point c , coordinate will be = \( \sqrt{k},0\) and point B, will be \(\sqrt{k},0\) Now the area of \(\triangle ABC = \frac{1}{8}\) or \(\frac{1}{2} * 2\sqrt{k}* k = \frac{1}{8}( base = \sqrt{k} + \sqrt{k}) = 2\sqrt{k}\) and height = point A and the distance = k) or \(k = \frac{1}{4} = 0.25\)
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Re: The figure shows the graph of the equation y = k – x^2, [#permalink]
06 Jan 2019, 18:46
pranab01 wrote: height = point A and the distance = k) How? Please explain.



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Re: The figure shows the graph of the equation y = k – x^2, [#permalink]
06 Jan 2019, 19:34
AE wrote: pranab01 wrote: height = point A and the distance = k) How? Please explain. Both xaxis and yaxis are perpendicular to each other, so the line from origin to A is the altitude of the triangle... And we know that when x=0, y=k, so coordinates of point A are (0,k).. Thus the height of the triangle is k
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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?



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Re: The figure shows the graph of the equation y = k – x^2, [#permalink]
07 Jan 2019, 06:00
chetan2u wrote: AE wrote: pranab01 wrote: height = point A and the distance = k) How? Please explain. Both xaxis and yaxis are perpendicular to each other, so the line from origin to A is the altitude of the triangle... And we know that when x=0, y=k, so coordinates of point A are (0,k).. Thus the height of the triangle is k Now I got It. My confusion was how the Point A is (0,k). Now it is clear. In equation if x=0, what is y, and here, y is k.



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Re: The figure shows the graph of the equation y =k–x^2, where k [#permalink]
04 Nov 2019, 21:22



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Re: The figure shows the graph of the equation y =k–x^2, where k [#permalink]
14 Dec 2019, 09:20
Merged similar topics. Up for further discussion
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Re: The figure shows the graph of the equation y =k–x^2, where k
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14 Dec 2019, 09:20





