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Re: the point (-1,0) lies on the parabola [#permalink]
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Explanation

Image

Here we have been given a parabola symmetric along x axis.

General Equation of parabola is : \(y^2=ax+b\) So rewrititng this equation in the form of quantities we can say:

\(x=\frac{1}{a}y^2 -\frac{b}{a}\)

Hence the two qty are equal when \(a=-1\) and \(\frac{b}{a}=-1\) or \(b=1\)

Now in order to calculate two variables namely \(a\) and \(b\) we need two eaquations. We can get one equation by putting the point (-1, 0) in the parabola equation \(y^2=ax+b\). Hence we cannot say if the answer is C or not!

Depending upon the second point of the parabola the quantities may or maynot be equal.

Hence D is the answer.
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Re: the point (-1,0) lies on the parabola [#permalink]
amorphous wrote:
NixonDutt wrote:
Someone please explain the question. This is one of the questions in test set.

The question is not complete. Put up a complete question.

I think, as the question is not complete, answer is nothing but D.
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Re: the point (-1,0) lies on the parabola [#permalink]
sandy wrote:
Explanation

Image

Here we have been given a parabola symmetric along x axis.

General Equation of parabola is : \(y^2=ax+b\) So rewrititng this equation in the form of quantities we can say:

\(x=\frac{1}{a}y^2 -\frac{b}{a}\)

Hence the two qty are equal when \(a=-1\) and \(\frac{b}{a}=-1\) or \(b=1\)

Now in order to calculate two variables namely \(a\) and \(b\) we need two eaquations. We can get one equation by putting the point (-1, 0) in the parabola equation \(y^2=ax+b\). Hence we cannot say if the answer is C or not!

Depending upon the second point of the parabola the quantities may or maynot be equal.

Hence D is the answer.



Hello,
a=-1 did you assume it because it is facing the left ?

Next, the parabola equation if it was sym. on the y axis is \(y^2=a(x-h)+K\) , while k and h are vertex points. Can we say for the parabola symmetric to x-axis that it is \(x^2=a(y-h)+K\)?
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Re: the point (-1,0) lies on the parabola [#permalink]
1
for left and right parabola we can use equation x = ay^2 +by +c as now we trace the roots on y -axis

we know vertex of parabola so we can use another equation of parabola x = (y - h)^2 + k
(h, k) = (-1,0)

thus (option A) x = (y + 1)^2 = y^2 + 2y + 1 and in option B given as -y^2 +1
so if we take y = 0 , we will get both equal and if we take y = 1 then A > B
option D
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Re: the point (-1,0) lies on the parabola [#permalink]
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