> 数学 >
点P是椭圆x^2/a^2+y^2/b^2=1任一点,O为坐标原点.(1)OP与X轴正半轴成α角求,模OP.
(2)Q为椭圆上另一点,并且OP⊥OQ,求证:1/(模OP^2) +1/(模QO^2)为定值.
人气:496 ℃ 时间:2020-02-06 00:41:11
解答
P(acosx,bsinx)
tanα=bsinx/acosx = b/a * tanx
tanx = a/b * tanα
tan^2 x = a^2/b^2 * tan^2 α
cos^2 x = 1/(a^2/b^2 * tan^2 α + 1)
= b^2 / (a^2tan^2 α + b^2)
sin^2 x = a^2tan^2 α / (a^2tan^2 α + b^2)
模OP = 根号(a^2(b^2 / (a^2tan^2 α + b^2)) + b^2(a^2tan^2 α / (a^2tan^2 α + b^2)))
= 根号(a^2b^2 / (a^2tan^2 α + b^2) + a^2b^2tan^2 α / (a^2tan^2 α + b^2)))
= 根号(a^2b^2(1+tan^2 α)/ (a^2tan^2 α + b^2))
= ab根号(1/(a^2tan^2α + b^2))/cosα
α'=α+pi/2
1/(模OP^2) +1/(模QO^2)
= cos^2α(a^2tan^2α + b^2)/a^2b^2 + cos^2α'(a^2tan^2α' + b^2)/a^2b^2
= (cos^2α(a^2tan^2α + b^2) + sin^2α(a^2cot^2α + b^2))/a^2b^2
= (a^2tan^2αcos^2α + b^2cos^2α + a^2sin^2αcot^2α + b^2sin^2α))/a^2b^2
= (a^2sin^2α + b^2cos^2α + a^2cos^2α + b^2sin^2α))/a^2b^2
= (a^2*1 + b^2*1)/a^2b^2
= (a^2 + b^2)/a^2b^2
所以1/(模OP^2) +1/(模QO^2)为定值
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版