Question 798328
You would expect exactly one intersection point.  Substitute for y and solve for x.


{{{4x^2+2(2x+3)^2=2}}}, that substitution done,
{{{2x^2+(2x+3)^2=1}}}
{{{2x^2+4x^2+12x+9=1}}}
{{{6x^2+12x+8=0}}}
{{{3x^2+6x+4=0}}}


{{{x=(-6+-sqrt(36-4*3*4))/(2*3)}}}
{{{x=(-6+-sqrt(36-48))/(2*3)}}}
{{{x=(-6+-sqrt(-12))/(2*3)}}}
{{{x=(-3+- i*sqrt(3))/(3)}}}, apparently NOT tangent!  A COMPLEX solution!


I've checked this over twice.  You would not be expecting a complex solution, but you want instead a single real solution.  One of your equations must be wrong.  



A better look at the graph:

{{{graph(300,300,-6,6,-6,6,-sqrt(-2x^2+1),sqrt(-2x^2+1),2x+3)}}}