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Re: A parabola in the coordinate geometry plane is represented by the equa [#permalink]
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vanam52923 wrote:
But i am not clear on statement 2.
It is given x>0 so we have to take positive x so quad 1 or 4th
but why cannot i take parabola as shown in my diagram case 2:
second parabola
its x intercept is positive and y is also positive.but it lies in 4th quad too(as shown in diagram)
Is it because then it wont b symmetric around y axis?
this is my confusion.
here
i am getting parabola should be symmetric to y and it should have +ve x intercept with positive y intercept so Case 2 second diagram fails but in case 2 diagram 1 ,i am not intercepting parabola any where on x axis.so how is this possible.
I read somewhere on net that
here for x intercept
y=0 if i put
i get x^2=-k
so x will have imaginary coordinates which is not possible so this parabola cannot have x intercept.Is this true.If yes then only Case 2 diagram 1 holds.Is it a thumb rule?
Thanking you in anticipation :)



You are given that the equation of the parabola is

y = x^2 + k
where k is POSITIVE (say k = 2). Then y = x^2 + 2

Since the co-efficient of x^2 is positive, it will be upward facing parabola.
x^2 + 2 will never be 0 (since x^2 cannot be negative) so it has no real roots. So it will NOT cut the x axis anywhere.
When x = 0, y = 2 so the parabola cuts the y axis at (0, 2).
Since there is no bx term, the parabola will be symmetric about the y axis (for both x = 3 and -3, y will be 11 etc)
Hence the parabola is going to lie above the x axis, symmetric about the y axis and intersecting the y axis at the point k (which we assumed 2 for ease).
Hence case 2 of your diagram is not possible. The parabola MUST look like case 1.

I hope this helps.
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Re: A parabola in the coordinate geometry plane is represented by the equa [#permalink]
VeritasKarishma wrote:
vanam52923 wrote:
But i am not clear on statement 2.
It is given x>0 so we have to take positive x so quad 1 or 4th
but why cannot i take parabola as shown in my diagram case 2:
second parabola
its x intercept is positive and y is also positive.but it lies in 4th quad too(as shown in diagram)
Is it because then it wont b symmetric around y axis?
this is my confusion.
here
i am getting parabola should be symmetric to y and it should have +ve x intercept with positive y intercept so Case 2 second diagram fails but in case 2 diagram 1 ,i am not intercepting parabola any where on x axis.so how is this possible.
I read somewhere on net that
here for x intercept
y=0 if i put
i get x^2=-k
so x will have imaginary coordinates which is not possible so this parabola cannot have x intercept.Is this true.If yes then only Case 2 diagram 1 holds.Is it a thumb rule?
Thanking you in anticipation :)



You are given that the equation of the parabola is

y = x^2 + k
where k is POSITIVE (say k = 2). Then y = x^2 + 2

Since the co-efficient of x^2 is positive, it will be upward facing parabola.
x^2 + 2 will never be 0 (since x^2 cannot be negative) so it has no real roots. So it will NOT cut the x axis anywhere.
When x = 0, y = 2 so the parabola cuts the y axis at (0, 2).
Since there is no bx term, the parabola will be symmetric about the y axis (for both x = 3 and -3, y will be 11 etc)
Hence the parabola is going to lie above the x axis, symmetric about the y axis and intersecting the y axis at the point k (which we assumed 2 for ease).
Hence case 2 of your diagram is not possible. The parabola MUST look like case 1.

I hope this helps.

Thank you so much mam.I am really indebted to u for all your help
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Re: A parabola in the coordinate geometry plane is represented by the equa [#permalink]
Bunuel wrote:
A parabola in the coordinate geometry plane is represented by the equation y = x^2 + k, where k is a constant greater than 0. Line L intersects this parabola at exactly one point. Is this point of intersection in Quadrant I?


(1) The slope of line L is positive.

(2) x is greater than 0.



VeritasKarishma Bunuel mikemcgarry

Is the answer option A or option D ?
Please clear our confusion...
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Re: A parabola in the coordinate geometry plane is represented by the equa [#permalink]
Does parabola equation comes in GMAT?
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Re: A parabola in the coordinate geometry plane is represented by the equa [#permalink]
Can someone please explain why st 1 is sufficient. If Line L's slope is +ve it can be anywhere i.e. it can intersect parabola in any quadrant. Isn't it ?
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Re: A parabola in the coordinate geometry plane is represented by the equa [#permalink]
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rsrighosh wrote:
Does parabola equation comes in GMAT?


Yes, it's covered in the GMAT syllabus
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Re: A parabola in the coordinate geometry plane is represented by the equa [#permalink]
OE
With a little number picking for x,y,
and k
or with a basic understanding of curved lines in the coordinate geometry plane, you can see that this parabola opens upward, is centered on the y axis, and has its lowest point (vertex) above the x axis (because k
is positive). The most important part of this problem is making sure that you have a proper visual representation of this parabola, so make sure that you take time to make a quick sketch before you start to analyze each statement.

If line L intersects this parabola at exactly one point, then visually you can see different possibilities along the parabola if you've drawn it correctly. The only possible intersection points are clearly in Quadrants I and II as the parabola does not even enter the other two quadrants. The question is asking whether that intersection point is in quadrant I or not.

Statement 1 tells you that line L has a positive slope. If you draw a few lines with positive slopes that only intersect the parabola at one point, you you can see that no matter the line, it must intersect the parabola in the 1st quadrant. If the slope is 0 (which is not allowed), then it would intersect at the y axis, and as the slope moves from 0 to infinity it can only intersect in the 1st quadrant. Statement 1 is sufficient. Eliminate (B), (C), and (E).

Statement 2 tells you that x
is positive which necessarily limits the intersection points to the 1st quadrant as all x
values in the 2nd quadrant are negative. Statement 2 is also sufficient and the answer is (D).
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Re: A parabola in the coordinate geometry plane is represented by the equa [#permalink]
In statement 1 we can play with numbers too, not only with intuition.
The slope of the tangent line to a curve is the value of the derivative of the function at that point. So let's derivate the function:
y' = 2x
which is positive only when x is positive.
And as y = x^2 + k takes only positive values, then the point is in the first quadrant.
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Re: A parabola in the coordinate geometry plane is represented by the equa [#permalink]
Pranjal3107 wrote:
Can someone please explain why st 1 is sufficient. If Line L's slope is +ve it can be anywhere i.e. it can intersect parabola in any quadrant. Isn't it ?


It can intersect in any quadrant but we have to make it satisfy one more condition in the question stem that the line intersects with parabola at exactly one point therefore that can happen only in Quadrant 1
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Re: A parabola in the coordinate geometry plane is represented by the equa [#permalink]
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Re: A parabola in the coordinate geometry plane is represented by the equa [#permalink]
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