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

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Math Expert
Joined: 02 Sep 2009
Posts: 58335
The figure shows the graph of the equation y = k – x^2, where k is a  [#permalink]

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28 Jan 2019, 00:56
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Difficulty:

55% (hard)

Question Stats:

60% (02:28) correct 40% (02:22) wrong based on 68 sessions

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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?

A. 1/4
B. 1/2
C. 1
D. 2
E. 4

Attachment:

#GREpracticequestion The figure shows the graph of the equation.jpg [ 11.01 KiB | Viewed 1234 times ]

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Re: The figure shows the graph of the equation y = k – x^2, where k is a  [#permalink]

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28 Jan 2019, 03:07
$$y = k - x^2$$

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
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Re: The figure shows the graph of the equation y = k – x^2, where k is a  [#permalink]

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28 Jan 2019, 03:23
1
Bunuel wrote:

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

Attachment:
#GREpracticequestion The figure shows the graph of the equation.jpg

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
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Joined: 21 May 2018
Posts: 10
Re: The figure shows the graph of the equation y = k – x^2, where k is a  [#permalink]

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30 Jan 2019, 19:17
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

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

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03 Feb 2019, 03:40
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
Math Expert
Joined: 02 Aug 2009
Posts: 7945
Re: The figure shows the graph of the equation y = k – x^2, where k is a  [#permalink]

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03 Feb 2019, 04:13
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

Hi..
You are absolutely correct with your approach
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Re: The figure shows the graph of the equation y = k – x^2, where k is a  [#permalink]

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03 Feb 2019, 04:42
chetan2u wrote:
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

Hi..
You are absolutely correct with your approach

Thank you for the confirmation!!
Makes me feel confident that I'm on a right track.
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Posts: 1682
Location: India
The figure shows the graph of the equation y = k – x^2, where k is a  [#permalink]

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19 Jul 2019, 04:16
Bunuel wrote:

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

Attachment:
#GREpracticequestion The figure shows the graph of the equation.jpg

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
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The figure shows the graph of the equation y = k – x^2, where k is a   [#permalink] 19 Jul 2019, 04:16
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