(1)由题意得:\left\{{\begin{array}{l}{\frac{c}{a}=\frac{{\sqrt{3}}}{2}}\\{\frac{3}{{{a^2}}}+\frac{1}{{4{b^2}}}=1}\end{array}}\right. 且a^{2}=b^{2}+c^{2} ,得\left\{{\begin{array}{l}{a=2}\\{b=1}\end{array}}\right. , 所以椭圆C 的方程为\frac{{{x^2}}}{4}+{y^2}=1 . (2) 证明:由椭圆方程可知,A\left(-2,0\right) ,B\left(2,0\right) , 设M(x_{1} ,y_{1}) ,则N(x_{1} ,-y_{1}) 且\frac{{x_1^2}}{4}+y_1^2=1 ; 则{k_1}=\frac{{{y_1}}}{{{x_1}+2}} ,{k_2}=\frac{{-{y_1}}}{{{x_1}-2}} ,则{k_1}⋅{k_2}=\frac{{-y_1^2}}{{x_1^2-4}}=\frac{{\frac{{x_1^2-4}}{4}}}{{x_1^2-4}}=\frac{1}{4} , 所以k_{1}\cdot k_{2} 为定值\frac{1}{4} .
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