sanjoo wrote:

The equation x = 2y^2 + 5y - 17, describes a parabola in the xy coordinate plane. If line l, with slope of 3, intersects the parabola in the upper-left quadrant at x = -5, the equation for l is

A. 3x + y + 15 = 0

B. y - 3x - 11 = 0

C. -3x + y - 16.5 = 0

D. -2x - y - 7 = 0

E. -3x + y + 13.5 = 0

The question is made to look difficult though it is pretty simple if you focus on just the line and use process of elimination. (Remember that GMAT does not focus on parabolas so basically, the question should be quite do-able even if someone doesn't know how to handle parabolas.)

We need equation of l. Its slope must be 3.

Slope in option A and D is not 3 so we are left with B, C and E

The line has a point (-5, y) on it where y is positive (since the point lies in upper left quadrant).

In options B and E, if you put x = -5, you get -ve value for y co-ordinate. So ignore them.

Answer must be (C)

You can certainly use the theoretical process of plugging x = -5 in the parabola equation, getting 2 values of y, taking the +ve value and hence finding the equation of the line using slope and a point:

y - y1 = m(x - x1)

But remember, it is a little time consuming and probably not what GMAT expects you to do. The options will be such that the calculation will not be tedious if you know what you are looking for.

Thanks alot.. Btw nice trick ..! i didnt think abt it ..its easy and not time consuming..!!