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13 May 2017, 02:26
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Difficulty:
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66% (01:15) correct
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At how many points do the two graphs named as "First Graph" and "Second Graph" intersect?
1) The Equation of the first graph is x^2 + y^2 = 4
2) The second graph has equation y = ax^2 - 5 where a <0
Source: https://www.GMATinsight.com
1) The Equation of the first graph is x^2 + y^2 = 4
2) The second graph has equation y = ax^2 - 5 where a <0
Source: https://www.GMATinsight.com
TheMechanic
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22 Aug 2017, 09:31
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Can some one post the solution to this question? I believe the 2nd graph is a parabola with a downward curve. How can we find the points of intersection when the value of "a" is not given?
22 Aug 2017, 14:03
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When r is a number, the equation x^2 + y^2 = r^2 is always a circle with radius r centered at the origin. So if we have the equation x^2 + y^2 = 4, that's a circle of radius 2 with its center at (0, 0). The lowest point on that circle is at (0, -2).
You don't really need to know about parabolas for GMAT questions, but y = ax^2 - 5 will be a parabola, and if a is negative, it will be a downwards parabola (like an upside-down U shape), with y-intercept at -5. So the highest point on that parabola is at (0, -5), and it can't meet the circle x^2 + y^2 = 4 anywhere.
You can see roughly why those things are true even if you don't know about equations for parabolas or circles. In this equation:
x^2 + y^2 = 4
if you think about how large or small y can be, since we're adding y^2 to a square (i.e. to something which is zero or greater) and getting 4 as a result, y^2 can't be larger than 4. So for any point (x, y) on this curve, it is always true that 2 > y > -2.
On this curve
y = ax^2 - 5
since a is negative, and x^2 is positive (or zero), it must be that ax^2 < 0, so ax^2 - 5 must be less than or equal to -5. Since y is equal to ax^2 - 5, the y coordinate of any point on this curve is never greater than -5.
So there can't be any points on both curves, because there is no value of y that can work in both equations.
You don't really need to know about parabolas for GMAT questions, but y = ax^2 - 5 will be a parabola, and if a is negative, it will be a downwards parabola (like an upside-down U shape), with y-intercept at -5. So the highest point on that parabola is at (0, -5), and it can't meet the circle x^2 + y^2 = 4 anywhere.
You can see roughly why those things are true even if you don't know about equations for parabolas or circles. In this equation:
x^2 + y^2 = 4
if you think about how large or small y can be, since we're adding y^2 to a square (i.e. to something which is zero or greater) and getting 4 as a result, y^2 can't be larger than 4. So for any point (x, y) on this curve, it is always true that 2 > y > -2.
On this curve
y = ax^2 - 5
since a is negative, and x^2 is positive (or zero), it must be that ax^2 < 0, so ax^2 - 5 must be less than or equal to -5. Since y is equal to ax^2 - 5, the y coordinate of any point on this curve is never greater than -5.
So there can't be any points on both curves, because there is no value of y that can work in both equations.
