How do you find the set of values for kk which the line x+3y=kx+3y=k intersects the curve y^2=2x+3y2=2x+3 at 2 real and distinct points?

1 Answer
Oct 4, 2015

I found k">"-6k>6

Explanation:

We can try taking the first equation and substitute for xx into the second:
x=k-3yx=k3y into the second:
y^2=2(k-3y)+3y2=2(k3y)+3
y^2=2k-6y+3y2=2k6y+3
Rearranging:
y^2+6y-(2k+3)=0y2+6y(2k+3)=0
We use the Quadratic Formula to solve for yy:
y_(1,2)=(-6+-sqrt(36+4(2k+3)))/2=y1,2=6±36+4(2k+3)2=
=(-6+-sqrt(36+8k+12))/2=(-6+-sqrt(48+8k))/2=6±36+8k+122=6±48+8k2
To be real and distinct we need the argument of the square root to be >0>0, so:
48+8k>048+8k>0
k">"-48/8k>488
k">"-6k>6