Solution:
the ellipse (x−x0)2a2+(y−y0)2b2=1;x0=0;y0=3;then(x−0)2a2+(y−3)2b2=1;c=2;b2=a2−c2=5;(x−0)29+(y−3)25=1;the point(0; 3+√5);(0−0)29+(3+5−3)25=1;1=1Answer:the point (0,3+√5) lies on the ellipsethe\; ellipse\;\;\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1;\\x_0=0;y_{0}=3;then\\\frac{(x-0)^2}{a^2}+\frac{(y-3)^2}{b^2}=1;\\c=2;b^2=a^2-c^2=5;\\\frac{(x-0)^2}{9}+\frac{(y-3)^2}{5}=1;the\; point (0;\;3+√5);\\\frac{(0-0)^2}{9}+\frac{(3+\sqrt5-3)^2}{5}=1;\\1=1\\Answer:\\the\; point\; (0,3+√5)\; lies\; on \;the \;ellipsetheellipsea2(x−x0)2+b2(y−y0)2=1;x0=0;y0=3;thena2(x−0)2+b2(y−3)2=1;c=2;b2=a2−c2=5;9(x−0)2+5(y−3)2=1;thepoint(0;3+√5);9(0−0)2+5(3+5−3)2=1;1=1Answer:thepoint(0,3+√5)liesontheellipse
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