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Do lines y = ax^2 + b and y = cx^2 + d cross?

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Do lines y = ax^2 + b and y = cx^2 + d cross? [#permalink] New post 10 Dec 2010, 08:41
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Do lines \(y = ax^2 + b\) and \(y = cx^2 + d\) cross?[/b]

(1) \(a = -c\)
(2) \(b \gt d\)

my Q in the spoiler
[Reveal] Spoiler:
In statement A its said a=-c, using the linear eq you can say if a1/a2 does not equal b1/b2, the lines do intersect, so isn't statement A alone sufficient?
[Reveal] Spoiler: OA

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Re: Intersection [#permalink] New post 10 Dec 2010, 08:46
Expert's post
Good discussion at: m24-12-formatting-error-73363.html

I think there is an answer on your question in my post. If anything remains unclear please ask.

My post from the earlier discussion of this question:

Do lines \(y = ax^2 + b\) and \(y = cx^2 + d\) cross?

1. \(a = -c\)
2. \(b \gt d\)

"I'd like to comment on this one:

First of all: equations given ARE NOT linear equations. We know that if two different lines do not cross each other they are parallel. How can we tell, based on the equations, whether the lines are parallel? We can check the slopes of these lines: parallel lines will have the same slopes. NOT that the slopes of lines must be negative reciprocals of each other (as it was mentioned in the earlier posts): in this case they are perpendicular not parallel.

Second of all: we have quadratic equations. These equations when drawn give parabolas. The question is: do they cross? This CAN NOT be transformed to the question: "are they parallel?" In the wast majority of cases the word "parallel" is used for the lines. Well we can say that concentric circles are parallel, BUT GMAT, as far as I know, uses this word ONLY about the lines (tutors may correct me if I'm wrong). Next, the word "parallel" when used for curves (lines, ...) means that these curves remain a constant distance apart. So strictly speaking two parabolas to be parallel they need not only not to intersect but also to remain constant distance apart. In this case, I must say that this can not happen. If a curve is parallel (as we defined) to the parabola it won't be quadratic: so curve parallel to a parabola is not a parabola. So I think that at this point we can stop considering this concept in regard to the original question.

So in which cases parabolas do not cross? There are number of possibilities: We can shift the vertex: the parabolas \(y=x^2\) and \(y=x^2+1\) will never intersect (note that they won't be exactly parallel but they will never intersect). We can consider downward and upward parabolas and in some cases they also never intersect... Of course there can be other cases as well.

As for the solution. We can follow the way dzyubam proposed (and I think it's the fastest way, provided we can identify correct examples) and consider two cases. First case: \(y=-x^2+1\) and \(y=x^2+0\) (upward and downward parabolas), which satisfies both statements, and see that in this case answer is YES, they cross each other; and the second case: \(y=x^2+1\) and \(y=-x^2+0\) (also upward and downward parabolas), which also satisfies both statements, and see that in this case answer is NO, they do not cross each other. Two different answers to the question, hence not sufficient.

Answer: E.

We can solve the question algebraically as well:

Do lines \(y = ax^2 + b\) and \(y = cx^2 + d\) cross?

(1) \(a = -c\) --> \(y_1= ax^2 + b\) and \(y_2=-ax^2 + d\), now if they cross then for some \(x\), \(ax^2+b=-ax^2 + d\) should be true --> which means that equation \(2ax^2+(b-d)=0\) must have a solution, some real root(s), or in other words discriminant of this quadratic equation must be \(\geq0\) --> \(d=0-8a(b-d)\geq0\)? --> \(d=-8a(b-d)\geq0\)? Now can we determine whether this is true? We know nothing about \(a\), \(b\), and \(d\), hence no. Not sufficient.

(2) \(b>d\) --> the same steps: if \(y_1= ax^2 + b\) and \(y_2= cx^2 + d\) cross then for some \(x\), \(ax^2 +b=cx^2+d\) should be true --> which means that equation \((a-c)x^2+(b-d)=0\) must have a solution or in other words discriminant of this quadratic equation must be \(\geq0\) --> \(d=0-4(a-c)(b-d)\geq0\)? --> \(d=-4(a-c)(b-d)\geq0\)? Now can we determine whether this is true? We know that \(b-d>0\) but what about \(a-c\)? Hence no. Not sufficient.

(1)+(2) \(a=-c\) and \(b>d\) --> \(y_1= ax^2 + b\) and \(y_2=-ax^2 + d\) --> same steps as above --> \(2ax^2+(b-d)=0\) --> and the same question remains: is \(d=-8a(b-d)\geq0\) true? \(b-d>0\) but what about \(a\)? Not sufficient.

Answer: E.

Hope it helps."
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Re: Intersection   [#permalink] 10 Dec 2010, 08:46
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