Coordinate plane

[BTGmoderatorRO]

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

[EconomistGMATTutor]

Hello Roland2rule.

Let's take a look. The equation of the parabola is $$x=2y^2+5y-17.$$ The equation of the line L is $$L:\ y-y_0=m\left(x-x_0\right)\Leftrightarrow\ \ y-y_0=3\left(x-x_0\right).$$ We just have to find a point that is on the line.

In order to find it we have to solve the quadratic equation $$-5=2y^2+5y-17\ \Leftrightarrow\ \ 2y^2+5y-12=0.$$ The Roots of this equation are given by $$y_{1,2}=\frac{-5\pm\sqrt{5^2-4\cdot2\cdot\left(-12\right)}}{2\cdot2}\ =\ \frac{-5\pm\sqrt{25+96}}{4}=\frac{-5\pm\sqrt{121}}{4}=\ \frac{-5\pm11}{4}.$$ The two solutions are y=6/4=3/2 and y=-16/4=-4.

Since the line intersects the parabola in the upper-left quadrant at x = -5, then we have to select the positive solution.

Therefore, the point of intersection is (-5,3/2). So, the equation of the line is $$L\ :\ y-\frac{3}{2}=3\left(x-\left(-5\right)\right)\ \Leftrightarrow\ \frac{2y-3}{2}=3x+15\ \Leftrightarrow\ 2y-3=6x+30$$ $$\ \Leftrightarrow\ \ -6x+2y-33=0\ \Leftrightarrow\ -3x+y-16.5=0.$$ Hence, the correct option is C.

I hope this explanation may help you.

Feel free to ask me again if you have a doubt.

Regards.