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The equation x = 2y^2 + 5y - 17, describes a parabola in the

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The equation x = 2y^2 + 5y - 17, describes a parabola in the xy coordinate plane. If line l, with slope of 3, intersects the parabola in the upper-left quadrant at x = -5, the equation for l is

A. 3x + y + 15 = 0
B. y - 3x - 11 = 0
C. -3x + y - 16.5 = 0
D. -2x - y - 7 = 0
E. -3x + y + 13.5 = 0
[Reveal] Spoiler: OA

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Let the equation of the line is y=mx+c given m=3 so equation of the line is y=3x+c

Given that the line intersects the parabola at (-5,y), then this is also a point on the parabola so substitute
x = 2y^2 + 5y - 17 intersects at -5,y

-5= 2y^2 + 5y - 17 ==> = 2y^2 + 5y - 12=0 ==> (y+4)(2y-3)=0

or y=-4 or y=3/2 but given they intersects the parabola in the upper-left quadrant so y>0 so y=3/2

so the point of intersection is (-5,3/2)

substitute this in the equation of the line to get c
y=3x+c

3/2=3*-5 +c

or c=16.5

so the equation of the line is y=3x+16.5

or -3x+y-16.5=0 a.k.a Option C
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IMO it is B
if the slope is 3 then y=mx + b
m must be 3 so I eliminate A and D.
Then Y in both the equal must be the same if I put X = -5.
B is right: y = 3x + 11
y = -15 +11 --> y = -4 --> then I sostitute -4 in the parabola's equation --> -5= 2 * -4^2 - 20 - 17
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Re: The equation x = 2y2 + 5y - 17, describes a parabola in the [#permalink]

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New post 17 Oct 2012, 13:36
IanSolo wrote:
IMO it is B
if the slope is 3 then y=mx + b
m must be 3 so I eliminate A and D.
Then Y in both the equal must be the same if I put X = -5.
B is right: y = 3x + 11
y = -15 +11 --> y = -4 --> then I sostitute -4 in the parabola's equation --> -5= 2 * -4^2 - 20 - 17



That is the trap in this question... Y has to be positive( intersects the parabola in the upper-left quadrant at x = -5)

So B is incorrect..
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Re: The equation x = 2y^2 + 5y - 17, describes a parabola in the [#permalink]

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New post 17 Oct 2012, 22:38
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sanjoo wrote:
The equation x = 2y^2 + 5y - 17, describes a parabola in the xy coordinate plane. If line l, with slope of 3, intersects the parabola in the upper-left quadrant at x = -5, the equation for l is

A. 3x + y + 15 = 0
B. y - 3x - 11 = 0
C. -3x + y - 16.5 = 0
D. -2x - y - 7 = 0
E. -3x + y + 13.5 = 0


The question is made to look difficult though it is pretty simple if you focus on just the line and use process of elimination. (Remember that GMAT does not focus on parabolas so basically, the question should be quite do-able even if someone doesn't know how to handle parabolas.)

We need equation of l. Its slope must be 3.

Slope in option A and D is not 3 so we are left with B, C and E

The line has a point (-5, y) on it where y is positive (since the point lies in upper left quadrant).
In options B and E, if you put x = -5, you get -ve value for y co-ordinate. So ignore them.

Answer must be (C)

You can certainly use the theoretical process of plugging x = -5 in the parabola equation, getting 2 values of y, taking the +ve value and hence finding the equation of the line using slope and a point:

y - y1 = m(x - x1)

But remember, it is a little time consuming and probably not what GMAT expects you to do. The options will be such that the calculation will not be tedious if you know what you are looking for.
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Re: The equation x = 2y^2 + 5y - 17, describes a parabola in the [#permalink]

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New post 18 Oct 2012, 10:13
VeritasPrepKarishma wrote:
sanjoo wrote:
The equation x = 2y^2 + 5y - 17, describes a parabola in the xy coordinate plane. If line l, with slope of 3, intersects the parabola in the upper-left quadrant at x = -5, the equation for l is

A. 3x + y + 15 = 0
B. y - 3x - 11 = 0
C. -3x + y - 16.5 = 0
D. -2x - y - 7 = 0
E. -3x + y + 13.5 = 0


The question is made to look difficult though it is pretty simple if you focus on just the line and use process of elimination. (Remember that GMAT does not focus on parabolas so basically, the question should be quite do-able even if someone doesn't know how to handle parabolas.)

We need equation of l. Its slope must be 3.

Slope in option A and D is not 3 so we are left with B, C and E

The line has a point (-5, y) on it where y is positive (since the point lies in upper left quadrant).
In options B and E, if you put x = -5, you get -ve value for y co-ordinate. So ignore them.

Answer must be (C)

You can certainly use the theoretical process of plugging x = -5 in the parabola equation, getting 2 values of y, taking the +ve value and hence finding the equation of the line using slope and a point:

y - y1 = m(x - x1)

But remember, it is a little time consuming and probably not what GMAT expects you to do. The options will be such that the calculation will not be tedious if you know what you are looking for.


+1 karishma..!!

Thanks alot.. Btw nice trick ..! i didnt think abt it ..its easy and not time consuming..!!
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Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

Senior Manager
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Joined: 06 Aug 2011
Posts: 399
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Kudos [?]: 212 [0], given: 82

Re: The equation x = 2y^2 + 5y - 17, describes a parabola in the [#permalink]

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New post 18 Oct 2012, 10:14
abhishekkpv wrote:
Let the equation of the line is y=mx+c given m=3 so equation of the line is y=3x+c

Given that the line intersects the parabola at (-5,y), then this is also a point on the parabola so substitute
x = 2y^2 + 5y - 17 intersects at -5,y

-5= 2y^2 + 5y - 17 ==> = 2y^2 + 5y - 12=0 ==> (y+4)(2y-3)=0

or y=-4 or y=3/2 but given they intersects the parabola in the upper-left quadrant so y>0 so y=3/2

so the point of intersection is (-5,3/2)

substitute this in the equation of the line to get c
y=3x+c

3/2=3*-5 +c

or c=16.5

so the equation of the line is y=3x+16.5

or -3x+y-16.5=0 a.k.a Option C



+1 Abhishek ..Thanks alot..!! :) ..
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Re: The equation x = 2y^2 + 5y - 17, describes a parabola in the   [#permalink] 18 Oct 2012, 10:14
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