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

Question

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.

(1) The slope of line L is positive.

(2) x is greater than 0.

BrentGMATPrepNow · Accepted Answer

Given : 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.

Target question :    Is the point of intersection in Quadrant I?

First off, here's what the graph of y = x² looks like: 
 https://i.imgur.com/MVYTSU6.png

Since we're adding   k   (which is positive) all of the y-coordinates of the graph of y = x² will be increase by   k   units. 
So the graph of y = x² +   k   will look something like this: 
 https://i.imgur.com/uSO46Ta.png

Important : Since the graph of y = x² +   k   lies in quadrants I and II only,   the point of intersection (of the line and parabola) must be in EITHER quadrant I OR quadrant II  )

Statement 1 : The slope of line L is positive 
This statement is sufficient. Here's why:
If a line of positive slope intersects the parabola in quadrant II, then that line will also intersect the parabola at a point in quadrant I.
For example: 
 https://i.imgur.com/QhxjxDP.png 
Since we're told line L intersects this parabola at exactly one point, we can be certain that  line L can't intersect the parabola in quadrant II 
This means  line L must intersect the parabola in quadrant I only 
Since we can answer the  target question  with certainty, statement 1 is SUFFICIENT

If you're not convinced, here's another way to look at it. 
If line L intersects this parabola at  exactly one point , then line L must be  tangent  to the parabola. 
So for example, line L might look like this: 
 https://i.imgur.com/BUzOVKu.png

Or like this: 
 https://i.imgur.com/RQj5Un8.png

As you can see, if line L has a positive slope  and  is tangent to the parabola,  the point of intersection must lie in quadrant I

Statement 2 : x is greater than 0 
I'm not crazy about this wording.
For clarity, I would write statement 2 as follows:  The x-coordinate of the point of intersection is positive  
This means line L must be tangent to the parabola at one of the   red dots   shown below
 https://i.imgur.com/PhM8B09.png 
Since all of the possible points of intersection are in quadrant I, we can be certain that  line L must intersect the parabola in quadrant I 
Since we can answer the  target question  with certainty, statement 2 is SUFFICIENT

Answer: D

Cheers,
Brent

PKN · Answer

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)

sumit411 · Answer

IMG_20180822_134818.jpg Imo A Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app

jhueglin · Answer

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.

jhueglin · Answer

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

rahul16singh28 · Answer

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.

PKN · Answer

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

rahul16singh28 · Answer

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.

PKN · Answer

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.

rahul16singh28 · Answer

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.

PKN · Answer

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.

rahul16singh28 · Answer

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.

Thanks  PKN .. I missed the Point that when Point of Intersection is Y-Axis (i.e. X = 0), the Slope will be Infinite.

PriyankaPalit7 · Answer

Hi experts,

chetan2u ,  Bunuel ,  VeritasKarishma ,  gmatbusters ,  amanvermagmat

Can you please provide your insights?

I feel statement 1 is insufficient.

Thanks,
Priyanka

GMATBusters · Answer

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

Hi experts,

chetan2u, Bunuel, VeritasKarishma, gmatbusters, amanvermagmat

Can you please provide your insights?

I feel statement 1 is insufficient.

Thanks,
Priyanka

Show Spoiler

Attachment:

WhatsApp Image 2018-12-08 at 23.09.28.jpeg

PriyankaPalit7 · Answer

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.

vanam52923 · Answer

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.
 
Please clear my doubts.

Bunuel   PKN   VeritasKarishma   gmatbusters 
 MathRevolution 
 chetan2u

JeffYin · Answer

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 · Answer

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.

I hope it was helpful.

vanam52923 · Answer

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

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

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