+86 The ellipse \(x^2+4y^2=4\) and the hyperbola \(x^2-m(y+2)^2 = 1\) are tangent. Compute \(m.\)
+9673 \(\begin{cases} x^2 + 4y^2 = 4\\ x^2 - m(y + 2)^2 = 1 \end{cases}\)
Substituting,
\(4 - 4y^2 - m(y + 2)^2 = 1\\ m(y + 2)^2 + 4y^2 - 3 = 0\\ (m + 4)y^2 + 4my + 4m - 3 = 0\)
Take \(\Delta = 0\), we have
\((4m)^2 - 4(m + 4)(4m - 3) = 0\\ 16m^2 - 4(4m^2 +13m -12) = 0\\ -52m+48 = 0\\ m = \dfrac{12}{13}\)
