求适合下列条件的椭圆标准方程:

(1) 与椭圆 $\frac{x^{2}}{2}+y^{2}=1$ 有相同的焦点, 且经过点 $\left(1, \frac{3}{2}\right)$

(2) 经过 $\mathrm{A}\left(2,-\frac{\sqrt{2}}{2}\right), \mathrm{B}\left(-\sqrt{2},-\frac{\sqrt{3}}{2}\right)$ 两点
(1) 粗圆 $rac{\mathrm{x}^{2}}{2}+\mathrm{y}^{2}=1$ 的焦点坐标为 $( \pm 1,0) ， ecause$ 椭圆过点 $\left(1, rac{3}{2}
ight)$, $	herefore 2 \mathrm{a}=\sqrt{(1+1)^{2}+\left(rac{3}{2}
ight)^{2}}+\sqrt{(1-1)^{2}+\left(rac{3}{2}
ight)^{2}}=4, \quad 	herefore \mathrm{a}=2, \quad \mathrm{~b}=\sqrt{3}, \quad 	herefore$ 方程为 $rac{\mathrm{x}^{2}}{4}+rac{\mathrm{y}^{2}}{3}=1$;(2) 设所求的椭圆方程为 $rac{x^{2}}{m}+rac{y^{2}}{n}=1, m>0, n>0, m 
eq n$. 把 $A\left(2,-rac{\sqrt{2}}{2}
ight), B\left(-\sqrt{2},-rac{\sqrt{3}}{2}
ight)$ 两点代入, 得: $\left\{egin{array}{l}rac{4}{m}+rac{rac{1}{2}}{n}=1  rac{2}{m}+rac{rac{3}{4}}{n}=1\end{array}
ight.$, 解得 $m=8, n=1, 	herefore$ 椭圆方程为 $rac{x^{2}}{8}+y^{2}=1$.