Some of the most challenging Quant questions on the GMAT involve Coordinate Geometry, so it’s important you have a solid grasp on the formulas and concepts on Test Day. You’ll see straight lines more than curved figures, but you may find it helpful to know the standard formula of the parabola in order to tackle some of the toughest Coordinate Geometry questions.
The standard equation for a parabola is y = ax2 + bx + c. Here c represents the y-intercept. Remember that a standard equation in which a variable is squared will never make a straight line – that x2 tells us we’re looking at a parabola. Let’s look at some questions!
1. In the xy plane, line P intersects y=x2 + 3 at how many points?
(1) Line P has a slope of 0
(2) Line P passes through (-2, 7)
It’s possible for line P to intersect y = x2 + 3 at 0, 1, or 2 points. For a statement to be sufficient, it will have to eliminate two out of those three possibilities.
We can plug in values for “x” for y = x2 + 3 to start to see how this parabola looks. (0, 3), (-1, 4) and (1, 4), for example, will be coordinate pairs on that parabola.
Statement (1) tells us that Line P is horizontal. Without more information, however, we cannot eliminate possibilities. Insufficient.
Statement (2) tells us one point on Line P, but we still don’t know its slope. It’s possible for Line P to intersect the parabola at 1 or 2 points.
Combined, the statements are sufficient. If Line P is horizontal, it can only intersect the parabola in two places.
2. If f(x) = ax2 + bx + c, and a ≠0, at what point does function f(x) intersect the y-axis?
(1) Function f(x) intersects the x-axis exactly twice, at (-6,0) and (-2,0).
(2) a = 2
This looks like a parabola question, but really it’s testing your knowledge of quadratics. Let’s start with statement (1). If (-6,0) and (-2,0) are both points on f(x), however we don’t know the shape of f(x). It could open upwards or downwards, which would affect the point it crosses the y-axis. Statement (2), tells us that a = 2. When a is positive, the parabola opens upwards. Therefore, the correct response is (C).
We could also solve this algebraically by plugging in x = -6 and x = -2 into f(x): a(-6)2 + b(-6) + c = 0 simplifies to 36a – 6b + c = 0.
a(-2)2 + b(-2) + c =0 simplifies to 4a – 2b + c = 0. We have two equations with three unknowns. Insufficient. Statement (2), however, provides us with one unknown. Now we’ll have two equations with two unknowns, and we’ll be able to solve.
Remember that parabolas are not extensively on the GMAT curriculum, but the GMAT will use them to test your knowledge of functions and quadratics.
Vivian Kerr is a regular contributor to the Veritas Prep blog, providing tips and tricks to help students better prepare for the GMAT and the SAT.