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A parabola in the coordinate geometry plane is represented by the equa

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

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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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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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New post 22 Aug 2018, 01:23
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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A parabola in the coordinate geometry plane is represented by the equa  [#permalink]

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New post 22 Aug 2018, 07:46
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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New post 22 Aug 2018, 08:07
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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A parabola in the coordinate geometry plane is represented by the equa  [#permalink]

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New post 22 Aug 2018, 10:04
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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New post 22 Aug 2018, 19:31
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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New post 22 Aug 2018, 21:27
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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Re: A parabola in the coordinate geometry plane is represented by the equa  [#permalink]

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New post 22 Aug 2018, 21:58
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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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A parabola in the coordinate geometry plane is represented by the equa  [#permalink]

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New post 23 Aug 2018, 00:21
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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A parabola in the coordinate geometry plane is represented by the equa  [#permalink]

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New post 23 Aug 2018, 00:41
1
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.

As per your comments,
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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Re: A parabola in the coordinate geometry plane is represented by the equa  [#permalink]

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New post 23 Aug 2018, 00:51
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.

As per your comments,
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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Re: A parabola in the coordinate geometry plane is represented by the equa  [#permalink]

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New post 23 Aug 2018, 06:47
Hi Bunuel,

Can you please provide your approach.

TIA

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Re: A parabola in the coordinate geometry plane is represented by the equa &nbs [#permalink] 23 Aug 2018, 06:47
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