以F1F2为焦点的椭圆x²/a²+y²/b²=1上一动点P,当角F1PF2最大时,角PF1F2的正切值为2,求离心率大小
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/11 04:43:56
![以F1F2为焦点的椭圆x²/a²+y²/b²=1上一动点P,当角F1PF2最大时,角PF1F2的正切值为2,求离心率大小](/uploads/image/z/3705053-5-3.jpg?t=%E4%BB%A5F1F2%E4%B8%BA%E7%84%A6%E7%82%B9%E7%9A%84%E6%A4%AD%E5%9C%86x%26%23178%3B%2Fa%26%23178%3B%2By%26%23178%3B%2Fb%26%23178%3B%3D1%E4%B8%8A%E4%B8%80%E5%8A%A8%E7%82%B9P%2C%E5%BD%93%E8%A7%92F1PF2%E6%9C%80%E5%A4%A7%E6%97%B6%2C%E8%A7%92PF1F2%E7%9A%84%E6%AD%A3%E5%88%87%E5%80%BC%E4%B8%BA2%2C%E6%B1%82%E7%A6%BB%E5%BF%83%E7%8E%87%E5%A4%A7%E5%B0%8F)
xRN@~*E61Y* o>^Dpz7(qD%$9ԐHG;D;^-:Ck$t3ŬܘDIΣM}aqZޞu?}:`tM=g2aC?hɅz"ގl
yW]owQ֠&_axɏdԎd#F5.gعx '
+,m2gCRw}jApO!2ĩ|O3NfT՞x:\SJ*
AY%)(so[ŎwPv[(*zΒU^:;yntޠ[Ek=;I#eXFp
以F1F2为焦点的椭圆x²/a²+y²/b²=1上一动点P,当角F1PF2最大时,角PF1F2的正切值为2,求离心率大小
以F1F2为焦点的椭圆x²/a²+y²/b²=1上一动点P,当角F1PF2最大时,角PF1F2的正切值为2,
求离心率大小
以F1F2为焦点的椭圆x²/a²+y²/b²=1上一动点P,当角F1PF2最大时,角PF1F2的正切值为2,求离心率大小
这种张角问题一般考虑用余弦定理.
cos∠F1PF2 = (F1P^2+F2P^2-F1F2^2)/(2F1P*F2P)
=[(F1P+F2P)^2-2F1P*F2P-F1F2^2])/(2F1P*F2P)
=-1+4(a^2-c^2)/(2F1P*F2P).1&
由于f(x) = cosx 在 x ∈(0,π)时是减函数,所以∠F1PF2 取最大值时cos∠F1PF2取最小值
由1&, cos∠F1PF2 ≤ -1+4(a^2-c^2)/[(F1P+F2P)^2/2]=-1+2(a^2-c^2)/a^2 当且仅当F1P=F2P时取等号,也就是说此时P是椭圆短轴顶点
所以 tan∠PF1F2=b/c=2
所以 e=c/a =1/√5=√5/5
0.25