2.(2020·新高考全国I卷)已知椭圆 :dfrac ({x)^2}({a)^2}+dfrac ({y)^2}({b)^2}=1(agt b-|||-gt 0) 的离心率为 dfrac (sqrt {2)}(2), 且过点A(2,1).-|||-(1)求C的方程;-|||-(2)点M,N在C上,且 bot AN bot MN, D为垂足.-|||-AN,-|||-证明:存在定点Q,使得|DQ|为定值.

16.-|||-已知椭圆 :dfrac ({x)^2}({a)^2}+dfrac ({y)^2}({b)^2}=1(agt bgt 0) 的离心率为 dfrac

(2024·新课标I卷)已知A(0,-|||-3)和 (3,dfrac (3)(2)) 为椭圆 :dfrac ({x)^2}({a)^2}+dfrac ({y)

6.已知A(0,3)和 (3,dfrac (3)(2)) 为椭圆 :dfrac ({x)^2}({a)^2}+dfrac ({y)^2}({b)^2}=1(ag

[题目]-|||-椭圆 dfrac ({x)^2}(3)+dfrac ({y)^2}(4)=1 的离心率是 __

(B) dfrac (x)(y)((y+1))^2-|||-(C) ^2((x+dfrac {1)(x))}^2. (D) dfrac (y)(x)((y+1)

设L为椭圆dfrac ({x)^2}(2)+dfrac ({y)^2}(3)=1，其周长为a，则dfrac ({x)^2}(2)+dfrac ({y)^2}(3

2.已知 dfrac (x)(x+y)=dfrac (1)(3) ,求 dfrac ({x)^2-(y)^2}(2xy+{y)^2} 的值.

2.求曲线 ^dfrac (2{3)}+(y)^dfrac (2{3)}=(a)^dfrac (2{3)} 在点 (dfrac (sqrt {2)}(4)a,d

2.求曲线 ^dfrac (2{3)}+(y)^dfrac (2{3)}=(a)^dfrac (2{3)} 在点 (dfrac (sqrt {2)}(4)a,d

. _(Y)(y)=dfrac (1)(2sqrt {2pi )}(e)^-dfrac (y{2)} ,gt 0-|||-bigcirc ._(Y)(y)=d