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Is there an intersection between the line (Y = aX - b) and the parabol

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Is there an intersection between the line (Y = aX - b) and the parabol  [#permalink]

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New post 18 Jul 2016, 07:28
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

44% (02:30) correct 56% (01:04) wrong based on 80 sessions

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Re: Is there an intersection between the line (Y = aX - b) and the parabol  [#permalink]

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New post 18 Jul 2016, 22:16
i am getting B

for
1) the parabola will be up y=+ve intercept and line will have y=-ve intercept.

So won't intersect.

2) opp as first.
line will intersect,


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Re: Is there an intersection between the line (Y = aX - b) and the parabol  [#permalink]

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New post 19 Jul 2016, 02:58
Bunuel wrote:
Is there an intersection between the line (Y = aX - b) and the parabola (Y = X^2 + b)?

(1) a < 0.
(2) 0 > b.


I am also getting B.

1) if a<0 => Slope of the line is negative. But it may happen that the line is drawn outside the parabola region. Also, when x=0, Parabola passes through (0,b) while the line passes through (0,-b). hence we are not sure if it will intersect or not. -- Not Sufficient.

2) if b<0= > Parabola intersect the y axis at a negative coordinate and the line intersect at the positive coordinate. Thus, whatever the slope of line is , it will always intersect the Parabola. -- Sufficient.

Ans -- B.
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Is there an intersection between the line (Y = aX - b) and the parabol  [#permalink]

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New post Updated on: 25 Jul 2016, 07:14
1
So here is my first attempt at posting a reply to any question. Help me if i go wrong :oops:

To figure out whether these two intersect, we can solve their equations.
Equating the RHS of both equations, we get :
\(ax - b = x^2 + b\)
\(x^2 - ax + 2b = 0\)

This is of the form\(A x^2 + B x + C = 0\)
So if the equation has real numbers as roots, we can conclude that the parabola and circle do intersect.

For the equation to have real roots, \(B^2 - 4AC\) should be a positive number, so that \(\sqrt{B^2-4AC}\) is real.

In our equation \(B^2 - 4AC = a^2 - 4*1*2b = a^2 - 8b\)
We don't know whether \(a^2> 8b\) from (1). Therefore we don't know if the equation has real roots.
From (2), we can conclude that since b<0 , -8b>0. Therefore, \(B^2-4AC\) is positive. Hence, the answer should be B.

Therefore, in my opinion the answer should be E.

Originally posted by apoorvajoshi on 25 Jul 2016, 06:06.
Last edited by apoorvajoshi on 25 Jul 2016, 07:14, edited 1 time in total.
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Re: Is there an intersection between the line (Y = aX - b) and the parabol  [#permalink]

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New post 25 Jul 2016, 07:09
apoorvajoshi wrote:
So here is my first attempt at posting a reply to any question. Help me if i go wrong :oops:

To figure out whether these two intersect, we can solve their equations.
Equating the RHS of both equations, we get :
\(ax - b = x^2 + b\)
\(x^2 - ax + 2b = 0\)

This is of the form\(A x^2 + B x + C = 0\)
So if the equation has real numbers as roots, we can conclude that the parabola and circle do intersect.

For the equation to have real roots, \(B^2 - 4AC\) should be a positive number, so that \(\sqrt{B^2-4AC}\) is real.

In our equation \(B^2 - 4AC = a^2 - 4*1*2b = a^2 - 8b\)
We don't know whether \(a^2> 8b\) from (1), (2) or (1) and (2). Therefore we don't know if the equation has real roots.

Therefore, in my opinion the answer should be E.


Apoorva, Why did you rule out Statement2?

Statement2 says that b<0, which means it is negative. And, \(a^2\) is definitely a non negative value.
Which would mean that \(a^2 - 8b\) is a positive value and the equation in question can have real roots.
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Re: Is there an intersection between the line (Y = aX - b) and the parabol  [#permalink]

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New post 25 Jul 2016, 07:12
winionhi wrote:
apoorvajoshi wrote:
So here is my first attempt at posting a reply to any question. Help me if i go wrong :oops:

To figure out whether these two intersect, we can solve their equations.
Equating the RHS of both equations, we get :
\(ax - b = x^2 + b\)
\(x^2 - ax + 2b = 0\)

This is of the form\(A x^2 + B x + C = 0\)
So if the equation has real numbers as roots, we can conclude that the parabola and circle do intersect.

For the equation to have real roots, \(B^2 - 4AC\) should be a positive number, so that \(\sqrt{B^2-4AC}\) is real.

In our equation \(B^2 - 4AC = a^2 - 4*1*2b = a^2 - 8b\)
We don't know whether \(a^2> 8b\) from (1), (2) or (1) and (2). Therefore we don't know if the equation has real roots.

Therefore, in my opinion the answer should be E.


Apoorva, Why did you rule out Statement2?

Statement2 says that b<0, which means it is negative. And, \(a^2\) is definitely a non negative value.
Which would mean that \(a^2 - 8b\) is a positive value and the equation in question can have real roots.



Yep totally missed that. Thank you :)
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Re: Is there an intersection between the line (Y = aX - b) and the parabol  [#permalink]

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New post 06 Oct 2016, 01:18
Bunuel wrote:
Is there an intersection between the line (Y = aX - b) and the parabola (Y = X^2 + b)?

(1) a < 0.
(2) 0 > b.


i thought this way.

since the min point of this parabola is (0,b) thus it is symmetrical around the y axis and since the co-oficient of x^2 is 1 then it is opened upwards and its x axis intercepts are dependent on b (+or- (-4b)^1/2. if b is -ve then it has 2 x intercepts and y intercept is -ve and and if +ve then it has no x intercept and y intercept is +ve.

in all cases ( regardless of slope) if the line y-intercept is > the y intercept of the parabola it ll always intersect it thus the solution to this question depends on (b)

from one no idea about b ...insuff

from 2 , b is -ve thus the y intercept of the parabola is -ve and that of the line is +ve thus they surely intersect.

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Re: Is there an intersection between the line (Y = aX - b) and the parabol  [#permalink]

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New post 20 Oct 2018, 12:48
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Re: Is there an intersection between the line (Y = aX - b) and the parabol &nbs [#permalink] 20 Oct 2018, 12:48
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