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

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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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New post 28 Jan 2019, 00:56
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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
#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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New post 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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New post 28 Jan 2019, 03:23
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Bunuel wrote:
Image
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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Re: The figure shows the graph of the equation y = k – x^2, where k is a  [#permalink]

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New post 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

Therefore answer = A
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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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New post 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
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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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New post 03 Feb 2019, 04:13
Mislead 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
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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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New post 03 Feb 2019, 04:42
chetan2u wrote:
Mislead 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


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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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New post 19 Jul 2019, 04:16
Bunuel wrote:
Image
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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