Parabolas and Triangles

[bretania]

Could someone please help? I don't know where I got it wrong but I am getting (d). The OA is (b). Could it be that the problem is open to 2 interpretations, hence 2 possible answers?


"In an X-Y plane, the three vertices of the triangle ABC are at the X and Y intercepts of the parabola defined by y = k - x2. If the base of the triangle lies along the X-axis and the area of the triangle is 64, what is k?"

(a) 20
(b) 16
(c) 12
(d) 8
(e) 4

[totoro]

It would help if you drew the figure and worked from there...

y-intercept=k
x-intercept=+-sqrt(k)

Since the area of the triangle is 64,
(k*2sqrt(k))/2 = 64

If you solve this equation correctly, you should get k=16.

[clock60]

Originally Posted by bretania

Could someone please help? I don't know where I got it wrong but I am getting (d). The OA is (b). Could it be that the problem is open to 2 interpretations, hence 2 possible answers?


"In an X-Y plane, the three vertices of the triangle ABC are at the X and Y intercepts of the parabola defined by y = k - x2. If the base of the triangle lies along the X-axis and the area of the triangle is 64, what is k?"

(a) 20
(b) 16
(c) 12
(d) 8
(e) 4

ok bretania let try i. can not imagine where did you find this problem. As for me it is awesome.
first, we deal with parabola and graph of this parabola tends down
as coefficient near x negative y=-x^2+k
then the top of parabola has coordinate (0,k)
after let us try point where graph intersect X line
y=0 and x^2=k so x= sq root from k
when we apply to answers we see that only one meaning of k is a perfect square and it is 16 so the answer is B
it certainly may be coincidence
and we check other way
the problem mentions triangle
the base of it lies on X line and it touches parabola
i do not know why but it seems to me that hight of triangle ends on the top
of parabola graph so the height =k
the area of triangle=1/2*base*height
so Area=1/2*base*k.
Parabola lies symmetricaly in relation to Y line , so the the base of triangle
is a distance where graph of parabola intersects X line
coordinate of the point we find above
y=0, x^2=k, k=16, x=4 so the base of the triangle is 8.
Area = 1/2*8(base)*16(height,k)=64 as in the problem.
I have no idea better just now and admit that solving is not perfect we must apply to gurus
but what it the source of problem

[rb999]

probably helps to draw it out and trial and error...