The ellipse x2+4y2=4 and the hyperbola x2−m(y+2)2=1 are tangent. Compute m.
{x2+4y2=4x2−m(y+2)2=1
Substituting,
4−4y2−m(y+2)2=1m(y+2)2+4y2−3=0(m+4)y2+4my+4m−3=0
Take Δ=0, we have
(4m)2−4(m+4)(4m−3)=016m2−4(4m2+13m−12)=0−52m+48=0m=1213