GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
The figure shows the graph of the equation y =k–x^2, where k
[#permalink]
29 Oct 2017, 01:27
29 Oct 2017, 01:27
5
1
12
Bookmarks
00:00
Question Stats:
55% (03:15) correct
44% (02:38) wrong based on 54 sessions
Hide Show timer Statistics
Attachment:
Coordinate Geometry - area- NE question.png [ 10.06 KiB | Viewed 29669 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
[#permalink]
31 Oct 2017, 00:55
31 Oct 2017, 00:55
5
1
Bookmarks
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)
[#permalink]
20 Sep 2018, 04:29
20 Sep 2018, 04:29
1
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
