The figure shows the graph of the equation y = k – x^2, where k is a

Question

https://gmatclub.com/forum/download/file.php?id=46176 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? A. 1/4 B. 1/2 C. 1 D. 2 E. 4 #GREpracticequestion The figure shows the graph of the equation.jpg

eswarchethu135 · Answer

y = k - x^2

Let point A be (0,a) and B be (b,0) and C be (-c,0). These three points lie on the curve. So these points can be substituted in the above equation.

a = k  ________(1)
 b^2 = k  ______(2)
 c^2 = k  ______(3)

Area of triangle ABC is  \frac{1}{8} 
 \frac{1}{2}*BC*AO = \frac{1}{8}  (O is the center of the coordinate system)

BC = b+c , AO = a

\frac{1}{2}*(b+c)*a = \frac{1}{8}

4a = b+c ___________(4)

Substitute eq (1) in eq (4)

4k = b+c

Now substitute eq (2) and eq (3) in the above equation.

b^2 = c^2 = k

b = c = \sqrt{k}

4k = \sqrt{k} + \sqrt{k}

4k = 2\sqrt{k}

2k = \sqrt{k}

Squaring on both sides,

4k^2 = k

k(4k - 1) = 0

k = 0 or  \frac{1}{4}

Since k cannot be zero, it should be  \frac{1}{4}

OPTION:  A

Archit3110 · Answer

good question

from given info we can determine points triangle a,b,c, 
a( 0,a); b( b,0) ; c ( -c,0)
using given eqn  y = k – x^2 
we can determine 
a= k, b^2=k & c^2 =k---- (1)

for triangle ABC 
we can say
0.5* BC* AO = 1/8
BC= oc+ob
upon solving
4a= b+c
substitute value (1)
4k= 2 sqrt k
2k= sqrtk
sqrt k = 1/2
square both sides
k = 1/4
IMO A

ak31 · Answer

y=K- x^{2} 
put x=0 to find out the coordinates of point A.
y=K
Therefore point A coordinates are (K,0)
now put y=0 to find out the coordinates of point B and C.
 x^{2} =k
x=+/- \sqrt{x}
Coordinate of B= ( \sqrt{K},0)
Coordinate of C= (- \sqrt{K},0)
Find the length of BC with length formula: \sqrt{ a^{2}  +  b^{2} }
Length of BC= \sqrt{ K^{2}  +  (-k)^{2} }
BC= 2\sqrt{K}
Area = 1/2 * AO * BC
1/8=1/2 * K * 2\sqrt{K}
k=1/4

Therefore answer = A

Mislead · Answer

can someone guide me if the below approach is correct or not:

y=k-x^2
if we take value of y=0 we have x^2=k.
we know Area of triange abc is 1/8 which:
1/2*(value of k)*(distance between BC) = 1/8

Using ans option:
A. 1/4===> if k=1/4 then x=+1/2 or x=-1/2 so the roots are (1/2,0) and (-1/2,0) so Distance between BC is 1 we know all the value: 1/2*1*1/4=1/8

ANS: A

Is this a correct way to solve this ?

Bunuel  and  chetan2u

chetan2u · Answer

Hi..
You are absolutely correct with your approach

Mislead · Answer

Thank you for the confirmation!! Makes me feel confident that I'm on a right track.

Kinshook · Answer

Height of triangle = Y-intercept = y=k when x=0
Base of the triangle/2 = x-intercept = x =  \sqrt{k}  when y=0
Area of triangle = 1/2 Base * Height =  k * \sqrt{k} = \frac{1}{8} 
k =  \frac{1}{4}

IMO A

Fdambro294 · Answer

The parabola is given by: y = -(x)^2 + k

the Y - Intercept of the Parabola, which is also Vertex A of the Triangle located at a Perpendicular straight line above the X-Axis, will be given when X = 0.

y = k

thus, the Height of the Triangle from Vertex A to the Base BC = K

Further, the 2 Roots (X-Intercepts) of the Parabola meet right at the Point of the 2 Vertices of the Triangle B and C which make up the Base along the X-Axis.

We can find these 2 X-Intercepts when Y = 0:

0 = -(x)^2 + k

(x)^2 = k

X = +sqrt(k)
and
X = -sqrt(k))

Thus, taking the Absolute Value of the Distance in units from the Origin to each Vertex B and C, the Base of the Triangle BC will be:

2 * sqrt(k)

Area of the Triangle = (1/2) * (Base) * (Height) =

(1/2) * (k) * (2*sqrt(k)) = 1/8

(k) * (sqrt(k)) = 1/8

the only A.C. that satisfies the above equation is answer choice A = 1/4

(1/4) * (sqrt(1/4)) = (1/4) * (1/2) = 1/8

(A)

adewale223 · Answer

4a = b + c please how?

Posted from my mobile device

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The figure shows the graph of the equation y = k – x^2, where k is a

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