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### HideShow timer Statistics 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.

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Imo A

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Re: A parabola in the coordinate geometry plane is represented by the equa  [#permalink]

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I think the problem is that the line intersects the parabola at exactly one point !!!
In conclusion we need a line that is parallel to the y axis and none of the two statements gives us tis information.
Based on your graphic the line will intersect the parabola twice.
Im not sure if I understood the question correct.
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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.

Given, quadratic equation: $$y=x^2+k$$, where k>0
Nature of the QE:-
a) Facing upward(a>0), above x-axis(c=k>0), symmetric about y-axis(x=-b/2a=0).

Intersection of line & Parabola:-
b) No of tangents from an external point:- 2, One tangent line has +ve slope & the other tangent has negative slope.
c) 3rd tangent to the parabola is the line y=k with zero slope.
d) The line x=0, intersects the parabola at exactly one point.slope=infinity
e) The point of intersection of line and parabola stays at Q1 when slope of the line is positive only.
N.B:- Tangent is also an intersection point.

Given, Line L intersects this parabola at exactly one point, which implies that either the tangent with +ve slope or the tangent with -ve slope or the tangent with is zero slope is considered.

St1:- The slope of line L is positive
Refer point(e), sufficient.

St2:- x is greater than 0
Since k is positive, hence y is also +ve. Therefore both x and y of the quadratic equation are +ve.
Hence, this portion represents the positive symmetry about y-axis.
Implies that slope of the line L is +ve.(Because we require exactly one intersection point in the right portion of y-axis of the parabola)
Refer point(e), sufficient.

Ans. (D)
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Re: A parabola in the coordinate geometry plane is represented by the equa  [#permalink]

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PKN wrote:
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.

Given, quadratic equation: $$y=x^2+k$$, where k>0
Nature of the QE:-
a) Facing upward(a>0), above x-axis(c=k>0), symmetric about y-axis(x=-b/2a=0).

Intersection of line & Parabola:-
b) No of tangents from an external point:- 2, One tangent line has +ve slope & the other tangent has negative slope.
c) 3rd tangent to the parabola is the line y=k with zero slope.
d) The line x=0, intersects the parabola at exactly one point.slope=infinity
e) The point of intersection of line and parabola stays at Q1 when slope of the line is positive only.
N.B:- Tangent is also an intersection point.

Given, Line L intersects this parabola at exactly one point, which implies that either the tangent with +ve slope or the tangent with -ve slope or the tangent with is zero slope is considered.

St1:- The slope of line L is positive
Refer point(e), sufficient.

St2:- x is greater than 0
Since k is positive, hence y is also +ve. Therefore both x and y of the quadratic equation are +ve.
Hence, this portion represents the positive symmetry about y-axis.
Implies that slope of the line L is +ve.(Because we require exactly one intersection point in the right portion of y-axis of the parabola)
Refer point(e), sufficient.

Ans. (D)

Thanks for that!
My mistake was that I forgot that tangents could also create intersection points.
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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.

I feel the Answer should be E.

Stmnt I: As the slope is +ve, the line can touch either the Vertex of Parabola or any other Single Point on Parabola in Ist Quadrant. When the line is touching the Vertex (i.e. on Y-Axis) the point of Intersection is Y-axis not in any quadrant. Hence, Insufficient.

Stmnt II: As x>0, the parabola lies only in first Quadrant. In this case also, Line L can Intersect the Origin of Parabola on Y - Axis or any other point on Parabola in 1st Quadrant. Hence, Insufficient.

Combining I & II, not sufficient.

Please correct me if I am wrong.
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Re: A parabola in the coordinate geometry plane is represented by the equa  [#permalink]

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rahul16singh28 wrote:
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.

I feel the Answer should be E.

Stmnt I: As the slope is +ve, the line can touch either the Vertex of Parabola or any other Single Point on Parabola in Ist Quadrant. When the line is touching the Vertex (i.e. on Y-Axis) the point of Intersection is Y-axis not in any quadrant. Hence, Insufficient.

Stmnt II: As x>0, the parabola lies only in first Quadrant. In this case also, Line L can Intersect the Origin of Parabola on Y - Axis or any other point on Parabola in 1st Quadrant. Hence, Insufficient.

Combining I & II, not sufficient.

Please correct me if I am wrong.

Note:- "Intersection or tangent on the axial point" doesn't lie on any quadrant.
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Re: A parabola in the coordinate geometry plane is represented by the equa  [#permalink]

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PKN wrote:
rahul16singh28 wrote:
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.

I feel the Answer should be E.

Stmnt I: As the slope is +ve, the line can touch either the Vertex of Parabola or any other Single Point on Parabola in Ist Quadrant. When the line is touching the Vertex (i.e. on Y-Axis) the point of Intersection is Y-axis not in any quadrant. Hence, Insufficient.

Stmnt II: As x>0, the parabola lies only in first Quadrant. In this case also, Line L can Intersect the Origin of Parabola on Y - Axis or any other point on Parabola in 1st Quadrant. Hence, Insufficient.

Combining I & II, not sufficient.

Please correct me if I am wrong.

Note:- "Intersection or tangent on the axial point" doesn't lie on any quadrant.

Hi PKN,

Didn't get you... !!

My point was the Intersection can be on the Axes or First Quadrant...!! When the Intersection is on Axes, then the point of Intersection doesn't lie on any Quadrant. So, we have both "Yes" & "No" to the question " Is this point of intersection in Quadrant I". So, it should be Insufficient.
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Hi PKN,

Didn't get you... !!

My point was the Intersection can be on the Axes or First Quadrant...!! When the Intersection is on Axes, then the point of Intersection doesn't lie on any Quadrant. So, we have both "Yes" & "No" to the question " Is this point of intersection in Quadrant I". So, it should be Insufficient.[/quote]

Hi rahul16singh28,

Statement 2:- x is greater than 0

According to you, there can be intersection points at the following instances :
1) On the y-axis
2) On the Q1
3) A vertical line on x-axis intersecting the parabola at exactly one point
4) On the x-axis

Let's analyze each of the above cases:
1) On the y-axis (means x=0), However, there is a caution , as per statement2 'x' is +ve only. Discard. N.B:- You know zero is neither +ve nor -ve.
We can't consider this case.

2) On the Q1:- This is obvious, you have explained it correctly.

3) A vertical line intersecting the parabola on Q1 passing through x-axis
We are given $$y=x^2+k$$
Also given, k>0
As per st2, x>0
Hence y>0
As x>0, y>0, hence the point of intersection lies in Q1.

4) On the x-axis: Since y=x^2+k, therefore, the parabola is above x-axis(k>0). So the line can never be a tangent to parabola on the x-axis. DISCARD

So, case-2 & 3 re the only valid scenarios in order to validate st2, which concludes that point of intersection lies in Q1.

Hope it helps.
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PKN wrote:
Hi PKN,

Didn't get you... !!

My point was the Intersection can be on the Axes or First Quadrant...!! When the Intersection is on Axes, then the point of Intersection doesn't lie on any Quadrant. So, we have both "Yes" & "No" to the question " Is this point of intersection in Quadrant I". So, it should be Insufficient.

Hi rahul16singh28,

Statement 2:- x is greater than 0

According to you, there can be intersection points at the following instances :
1) On the y-axis
2) On the Q1
3) A vertical line on x-axis intersecting the parabola at exactly one point
4) On the x-axis

Let's analyze each of the above cases:
1) On the y-axis (means x=0), However, there is a caution , as per statement2 'x' is +ve only. Discard. N.B:- You know zero is neither +ve nor -ve.
We can't consider this case.

2) On the Q1:- This is obvious, you have explained it correctly.

3) A vertical line intersecting the parabola on Q1 passing through x-axis
We are given $$y=x^2+k$$
Also given, k>0
As per st2, x>0
Hence y>0
As x>0, y>0, hence the point of intersection lies in Q1.

4) On the x-axis: Since y=x^2+k, therefore, the parabola is above x-axis(k>0). So the line can never be a tangent to parabola on the x-axis. DISCARD

So, case-2 & 3 re the only valid scenarios in order to validate st2, which concludes that point of intersection lies in Q1.

Hope it helps.[/quote]

Hi PKN,

Kudos for brief solution. My doubt is more on Statement I, as per Statement I, X = 0 or X > 0. So I feel, this is Insufficient.

Statement II is sufficient because X > 0, so the Intersection point cannot be on Y-axis.
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rahul16singh28 wrote:
PKN wrote:
Hi PKN,

Didn't get you... !!

My point was the Intersection can be on the Axes or First Quadrant...!! When the Intersection is on Axes, then the point of Intersection doesn't lie on any Quadrant. So, we have both "Yes" & "No" to the question " Is this point of intersection in Quadrant I". So, it should be Insufficient.

Hi rahul16singh28,

Statement 2:- x is greater than 0

According to you, there can be intersection points at the following instances :
1) On the y-axis
2) On the Q1
3) A vertical line on x-axis intersecting the parabola at exactly one point
4) On the x-axis

Let's analyze each of the above cases:
1) On the y-axis (means x=0), However, there is a caution , as per statement2 'x' is +ve only. Discard. N.B:- You know zero is neither +ve nor -ve.
We can't consider this case.

2) On the Q1:- This is obvious, you have explained it correctly.

3) A vertical line intersecting the parabola on Q1 passing through x-axis
We are given $$y=x^2+k$$
Also given, k>0
As per st2, x>0
Hence y>0
As x>0, y>0, hence the point of intersection lies in Q1.

4) On the x-axis: Since y=x^2+k, therefore, the parabola is above x-axis(k>0). So the line can never be a tangent to parabola on the x-axis. DISCARD

So, case-2 & 3 re the only valid scenarios in order to validate st2, which concludes that point of intersection lies in Q1.

Hope it helps.

Hi PKN,

Kudos for brief solution. My doubt is more on Statement I, as per Statement I, X = 0 or X > 0. So I feel, this is Insufficient.

Statement II is sufficient because X > 0, so the Intersection point cannot be on Y-axis.[/quote]

Hi rahul16singh28,

Statement-1:- The slope of line L is positive.
Implies that, If slope is denoted as m, then m>0
You know slope of form of the equation of a straight line: y=mx+c, where m is the slope and c is the y-intercept.

1) x=0, ok. if x=0, then y=m*0+c Or y=c implies slope is zero contradicts st1. DISCARD

N.B:- Slope is positive means Rise/Run>0 or vertical change in x-position/Horizontal change in y-position>0. We can't say x is positive or negative.
Example, points A(-5,-2) & B(2,3)
Slope=$$\frac{3-(-2)}{2-(-5)}=\frac{5}{7}>0$$
Here x=-5,2.
My point is: position of x is independent of slope when we determine x<0 or x>0 .

Refer my original explanation, all other possibilities will be rejected except point(e) when slope is positive.

Hope it helps.
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PKN wrote:
rahul16singh28 wrote:
PKN wrote:
Hi PKN,

Didn't get you... !!

My point was the Intersection can be on the Axes or First Quadrant...!! When the Intersection is on Axes, then the point of Intersection doesn't lie on any Quadrant. So, we have both "Yes" & "No" to the question " Is this point of intersection in Quadrant I". So, it should be Insufficient.

Hi rahul16singh28,

Statement 2:- x is greater than 0

According to you, there can be intersection points at the following instances :
1) On the y-axis
2) On the Q1
3) A vertical line on x-axis intersecting the parabola at exactly one point
4) On the x-axis

Let's analyze each of the above cases:
1) On the y-axis (means x=0), However, there is a caution , as per statement2 'x' is +ve only. Discard. N.B:- You know zero is neither +ve nor -ve.
We can't consider this case.

2) On the Q1:- This is obvious, you have explained it correctly.

3) A vertical line intersecting the parabola on Q1 passing through x-axis
We are given $$y=x^2+k$$
Also given, k>0
As per st2, x>0
Hence y>0
As x>0, y>0, hence the point of intersection lies in Q1.

4) On the x-axis: Since y=x^2+k, therefore, the parabola is above x-axis(k>0). So the line can never be a tangent to parabola on the x-axis. DISCARD

So, case-2 & 3 re the only valid scenarios in order to validate st2, which concludes that point of intersection lies in Q1.

Hope it helps.

Hi PKN,

Kudos for brief solution. My doubt is more on Statement I, as per Statement I, X = 0 or X > 0. So I feel, this is Insufficient.

Statement II is sufficient because X > 0, so the Intersection point cannot be on Y-axis.

Hi rahul16singh28,

Statement-1:- The slope of line L is positive.
Implies that, If slope is denoted as m, then m>0
You know slope of form of the equation of a straight line: y=mx+c, where m is the slope and c is the y-intercept.

1) x=0, ok. if x=0, then y=m*0+c Or y=c implies slope is zero contradicts st1. DISCARD

N.B:- Slope is positive means Rise/Run>0 or vertical change in x-position/Horizontal change in y-position>0. We can't say x is positive or negative.
Example, points A(-5,-2) & B(2,3)
Slope=$$\frac{3-(-2)}{2-(-5)}=\frac{5}{7}>0$$
Here x=-5,2.
My point is: position of x is independent of slope when we determine x<0 or x>0 .

Refer my original explanation, all other possibilities will be rejected except point(e) when slope is positive.

Hope it helps.[/quote]

Thanks PKN.. I missed the Point that when Point of Intersection is Y-Axis (i.e. X = 0), the Slope will be Infinite.
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GMAT 1: 640 Q45 V35 GMAT 2: 670 Q45 V37 Re: A parabola in the coordinate geometry plane is represented by the equa  [#permalink]

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Hi experts,

I feel statement 1 is insufficient.

Thanks,
Priyanka
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Hi PriyankaPalit7

If you visualize it, you will come to know that if the Line L has positive slope and it is tangent (intersects at a single point) to the parabola, it will intersect only in first Quadrant.

See the sketch. PriyankaPalit7 wrote:
Hi experts,

I feel statement 1 is insufficient.

Thanks,
Priyanka

Attachment: WhatsApp Image 2018-12-08 at 23.09.28.jpeg [ 48.64 KiB | Viewed 2398 times ]

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gmatbusters wrote:
Hi PriyankaPalit7

If you visualize it, you will come to know that if the Line L has positive slope and it is tangent (intersects at a single point) to the parabola, it will intersect only in first Quadrant.

See the sketch. PriyankaPalit7 wrote:
Hi experts,

I feel statement 1 is insufficient.

Thanks,
Priyanka

Attachment:
WhatsApp Image 2018-12-08 at 23.09.28.jpeg

Thanks I got it now. All other lines except L with positive slopes intersect the parabola at 2 points. So if the line intersects at 1 point and also has a positive slope, it is in the 1st quadrant.
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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.

Please clear my doubts on parabola:

here if we compare with ax^2+bx+c
we get b=0 there x=-b/2a =0 so x=0
i.e. axis of symmetry is y axis.

Now y intercept is when x=0
so y=k a positive number so parabola will face upwards

so why cannot parabola be like :
see figure case 1.
in that case line KL has positive slope but it cuts parabola at one point only in second quadrant.Why is this case invalid?
Do we assume OB(in figure)extends infinitely upwards so at to meet KL somewhere.

2)x>0
which x is this?
this is x intercept right .
not x of x=-b/2a
so to find this we need to put y=0
so x^2=-k
i.e x=+/- k
since x=+ve so
x=k
as k positive x is positive.

so how can that parabola be drawn which is symmetric to y but cuts x at positive points.
See figure Case 2 and case 3 in pic.

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Re: A parabola in the coordinate geometry plane is represented by the equa  [#permalink]

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vanam52923 wrote:
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.

Please clear my doubts on parabola:

here if we compare with ax^2+bx+c
we get b=0 there x=-b/2a =0 so x=0
i.e. axis of symmetry is y axis.

Now y intercept is when x=0
so y=k a positive number so parabola will face upwards

so why cannot parabola be like :
see figure case 1.
in that case line KL has positive slope but it cuts parabola at one point only in second quadrant.Why is this case invalid?
Do we assume OB(in figure)extends infinitely upwards so at to meet KL somewhere.

2)x>0
which x is this?
this is x intercept right .
not x of x=-b/2a
so to find this we need to put y=0
so x^2=-k
i.e x=+/- k
since x=+ve so
x=k
as k positive x is positive.

so how can that parabola be drawn which is symmetric to y but cuts x at positive points.
See figure Case 2 and case 3 in pic.

MathRevolution
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Regarding your doubt on statement 1: It is true that the graph of this parabola (including the part that you refer to as OB) extends infinitely upward (we don't have to assume this; it is definitely true). Because the slope of the parabola keeps increasing as you move up and to the right, its slope will eventually exceed the slope of line L, and line L will thus intersect the parabola at a second point. This is why the case that you drew isn't consistent with statement 1, and line L must be tangent to the parabola. In order for the slope of line L to be positive, it must be tangent to the parabola at a point where the slope of the parabola is positive, and the slope of the parabola is only positive in quadrant I.

Regarding your doubt on statement 2: The issue is the interpretation of "x is greater than 0". In this case, it actually means that we can only consider points where x is positive, in other words all of the points in quadrants I and IV. Since no part of the graph of y = x^2 + k is in quadrant IV, the point of intersection must be in quadrant I, so statement 2 is sufficient.

Please let me know if you have more questions!
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vanam52923
1. In the first figure that you have drawn the line will intersect parabola in two points. Parabola is not a finite curve. The question says it has only to intersect in one point.
2. X > 0 means in equation of parabola assume x>0 so entire parabola will be in first quadrant.

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A parabola in the coordinate geometry plane is represented by the equa  [#permalink]

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JeffYin wrote:
vanam52923 wrote:
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.

Please clear my doubts on parabola:

here if we compare with ax^2+bx+c
we get b=0 there x=-b/2a =0 so x=0
i.e. axis of symmetry is y axis.

Now y intercept is when x=0
so y=k a positive number so parabola will face upwards

so why cannot parabola be like :
see figure case 1.
in that case line KL has positive slope but it cuts parabola at one point only in second quadrant.Why is this case invalid?
Do we assume OB(in figure)extends infinitely upwards so at to meet KL somewhere.

2)x>0
which x is this?
this is x intercept right .
not x of x=-b/2a
so to find this we need to put y=0
so x^2=-k
i.e x=+/- k
since x=+ve so
x=k
as k positive x is positive.

so how can that parabola be drawn which is symmetric to y but cuts x at positive points.
See figure Case 2 and case 3 in pic.

MathRevolution
chetan2u

Regarding your doubt on statement 1: It is true that the graph of this parabola (including the part that you refer to as OB) extends infinitely upward (we don't have to assume this; it is definitely true). Because the slope of the parabola keeps increasing as you move up and to the right, its slope will eventually exceed the slope of line L, and line L will thus intersect the parabola at a second point. This is why the case that you drew isn't consistent with statement 1, and line L must be tangent to the parabola. In order for the slope of line L to be positive, it must be tangent to the parabola at a point where the slope of the parabola is positive, and the slope of the parabola is only positive in quadrant I.

Regarding your doubt on statement 2: The issue is the interpretation of "x is greater than 0". In this case, it actually means that we can only consider points where x is positive, in other words all of the points in quadrants I and IV. Since no part of the graph of y = x^2 + k is in quadrant IV, the point of intersection must be in quadrant I, so statement 2 is sufficient.

Please let me know if you have more questions!

JeffYin
thankyou so much for clearing it.
I am very much clear about statement 1 now.
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 Originally posted by vanam52923 on 09 Dec 2018, 01:41.
Last edited by vanam52923 on 09 Dec 2018, 01:43, edited 1 time in total.
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A parabola in the coordinate geometry plane is represented by the equa  [#permalink]

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ruchik wrote:
vanam52923
1. In the first figure that you have drawn the line will intersect parabola in two points. Parabola is not a finite curve. The question says it has only to intersect in one point.
2. X > 0 means in equation of parabola assume x>0 so entire parabola will be in first quadrant.

JeffYin wrote:
vanam52923 wrote:
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.

Please clear my doubts on parabola:

here if we compare with ax^2+bx+c
we get b=0 there x=-b/2a =0 so x=0
i.e. axis of symmetry is y axis.

Now y intercept is when x=0
so y=k a positive number so parabola will face upwards

so why cannot parabola be like :
see figure case 1.
in that case line KL has positive slope but it cuts parabola at one point only in second quadrant.Why is this case invalid?
Do we assume OB(in figure)extends infinitely upwards so at to meet KL somewhere.

2)x>0
which x is this?
this is x intercept right .
not x of x=-b/2a
so to find this we need to put y=0
so x^2=-k
i.e x=+/- k
since x=+ve so
x=k
as k positive x is positive.

so how can that parabola be drawn which is symmetric to y but cuts x at positive points.
See figure Case 2 and case 3 in pic.

MathRevolution
chetan2u

Regarding your doubt on statement 1: It is true that the graph of this parabola (including the part that you refer to as OB) extends infinitely upward (we don't have to assume this; it is definitely true). Because the slope of the parabola keeps increasing as you move up and to the right, its slope will eventually exceed the slope of line L, and line L will thus intersect the parabola at a second point. This is why the case that you drew isn't consistent with statement 1, and line L must be tangent to the parabola. In order for the slope of line L to be positive, it must be tangent to the parabola at a point where the slope of the parabola is positive, and the slope of the parabola is only positive in quadrant I.

Regarding your doubt on statement 2: The issue is the interpretation of "x is greater than 0". In this case, it actually means that we can only consider points where x is positive, in other words all of the points in quadrants I and IV. Since no part of the graph of y = x^2 + k is in quadrant IV, the point of intersection must be in quadrant I, so statement 2 is sufficient.

Please let me know if you have more questions!

ruchik
thankyou so much for clearing it.
I am very much clear about statement 1 now.
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 Originally posted by vanam52923 on 09 Dec 2018, 01:42.
Last edited by vanam52923 on 09 Dec 2018, 01:48, edited 1 time in total. A parabola in the coordinate geometry plane is represented by the equa   [#permalink] 09 Dec 2018, 01:42

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