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The figure shows the graph of the equation y =k–x^2, where k
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29 Oct 2017, 01:27
29 Oct 2017, 01:27
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Question Stats:
55% (03:15) correct
44% (02:38) wrong based on 54 sessions
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Coordinate Geometry - area- NE question.png [ 10.06 KiB | Viewed 29815 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
Show: :: OA
0.25
Re: The figure shows the graph of the equation y =k–x^2, where k
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31 Oct 2017, 00:55
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.
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.
General Discussion
Re: GRE Math Challenge #2 (170 level question)
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20 Sep 2018, 04:29
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=k-0 => 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=k-x^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
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=k-0 => 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=k-x^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
