這種題不難,但計(jì)算時(shí)要注意技巧
短軸 = 2b =4→b=2
焦點(diǎn)在x軸上,對(duì)稱軸為兩坐標(biāo)軸 那么設(shè)橢圓標(biāo)準(zhǔn)方程為 x^2/a^2+y^2/4=1
即 4x^2+a^2*y^2 = 4a^2 將 x = 4-2y代入.(簡(jiǎn)化計(jì)算量,用y=(4-x)/2帶入很繁瑣)
4(2(2-y))^2+a^2y^2 = 4a^2
→(16+a^2)y^2-64y+4(16-a^2) =0
又弦長(zhǎng) = √[(x1-x2)^2+(y1-y2)^2]=√[4(y1-y2)^2+(y1-y2)^2=|y1-y2| *√ 5=2√ 5
即 (y1-y2)^2 =4 |y1-y2|=2
(y1-y2)^2 =(y1+y2)^2-4y1y2=[64^2-16(16+a^2)(16-a^2)]/(16+a^2)^2
=16[4*64-256-a^4]/(16+a^2) ^2 [千萬不要把64^2乘出來,不然很麻煩.這是計(jì)算技巧]
→16a^4/(16+a^2)^2 = 4
→a^2/((16+a^2) = 2 →a^2 =16
所以,最后答案為 x^2/16+y^2/4=1