设f1 f2分别是椭圆x^2/a^2+y^2/b^ a>b>0 左右焦点 过f1 倾斜角为45度直线与椭圆交于pq两点且pq长度=4/3a
求离心率
设m 0 -1 满足 MP长度 =MQ长度 求其椭圆方程
由题意建立直角坐标系如图,已知f1(-c,0),m(0,-1),直线pq倾斜角为π/4,据此建立极坐标系如图。
以左焦点f1(-c,0)为极点o,以∠m f1 o=θ,设p点为(ρ1,π/4),q点为(ρ1,5π/4),m点为(ρ3,2π-θ).
由圆锥曲线统一的极坐标方程ρ=ep/(1-ecosθ)得
ρ1=ep/(1-ecosπ/4)=ep/(1-√2e/2)
ρ2=ep/(1-ecos5π/4)=ep/[1-ecos(π+π/4)]= ep/(1+ecosπ/4)=ep/(1+√2e/2)
ρ1+ρ2=ep/(1-√2e/2)+ep/(1+√2e/2)= ep(1+√2e/2+1-√2e/2)/(1-e^2/2)
= 2ep/(1-e^2/2)
= 4ep/(2-e^2)
由圆锥曲线统一方程定义可知P为焦准距,P=b^2/c,而 e为离心率,e=c/a ,
上式化简得ρ1+ρ2=4(c/a)(b^2/c)/(2-c^2/a^2)=4(b^2/a)/[(2a^2-c^2)/(a^2)]
=4ab^2/(a^2+b^2)
已知∣pq∣=ρ1+ρ2=4a/3
所以4ab^2/(a^2+b^2)= 4a/3,即3b^2=a^2+b^2, 2b^2=a^2 ……………………①
另根据余弦定理,△pf1m中∣pf1∣=ρ1,∣mf1∣=ρ3,
∣pm∣^2=∣pf1∣^2+∣mf1∣^2 - 2∣pf1∣∣mf1∣cos(θ+π/4)
∣pm∣^2=ρ1^2+ρ3^2 - 2ρ1ρ3 cos(θ+π/4)
△qf1m中∣qf1∣=ρ2,∣mf1∣=ρ3,
∣qm∣^2=∣qf1∣^2+∣mf1∣^2 - 2∣qf1∣∣mf1∣cos(θ+π/4)
∣qm∣^2=ρ2^2+ρ3^2 - 2ρ2ρ3 cos[π-(θ+π/4)]
∣qm∣^2=ρ2^2+ρ3^2 + 2ρ2ρ3 cos(θ+π/4)
已知∣pm∣=∣qm∣
所以ρ1^2+ρ3^2 - 2ρ1ρ3 cos(θ+π/4)= ρ2^2+ρ3^2 + 2ρ2ρ3 cos(θ+π/4)
即ρ1^2-ρ2^2= 2ρ2ρ3 cos(θ+π/4) + 2ρ1ρ3 cos(θ+π/4)
(ρ1-ρ2)(ρ1+ρ2)=2(ρ1+ρ2)ρ3 cos(θ+π/4)
ρ1-ρ2=2ρ3 cos(θ+π/4)
=2ρ3 (cosθcosπ/4-sinθsinπ/4)
=√2ρ3(cosθ-sinθ) ………………②
回到直角坐标系,∠θ的三角函数关系如图示,
cosθ= c/(√c^2+1)
sinθ= 1/(√c^2+1)
cosθ- sinθ=(c-1)/(√c^2+1)
ρ3=√c^2+1
将ρ1,ρ2以及上述cosθ,sinθ,ρ3代入②
②式左边如下
ep/(1-√2e/2)- ep/(1+√2e/2)=ep[1+√2e/2-(1-√2e/2)]/ (1-e^2/2)
= √2pe^2/(1-e^2/2)
= √2(b^2/c)(c^2/a^2)/(1-c^2/2a^2)
= (√2 c b^2/a^2)[2a^2/(2a^2-c^2)]
=2√2 c b^2/(a^2+b^2)………………③
②式右边如下
√2ρ3(cosθ-sinθ)= √2(√c^2+1)(c-1)/(√c^2+1)= √2(c-1) ………………④
由①③④得2√2 c b^2/(a^2+b^2)=√2c a^2/(3a^2/2)= 2√2c/3=√2c-√2
解得c=3,
a^2-b^2=c^2, 2b^2-b^2=c^2
b^2=c^2,b=3
a=√2b=3√2由此可知直角坐标系下该椭圆方程为x^2/18+y^2/9=1
答案不一定对,希望有所帮助。
会的话说啊