英文数学题一道+函数一道
来源:百度知道 编辑:UC知道 时间:2024/05/28 01:20:07
Show that the ellipse where the sum of the distances (1,0) and (0,1) is 二倍根二 can be written as
(x^2)/(a^2) + (y^2)/(b^2) = 1
where a and be are constants to be determined.
2. 求所有满足条件的函数
a) f(x+y) + f(x-y) = 2(x^2) + 2(y^2)
b) g(x+t) - g(x-t) = 4xt
第一题的第二个点是(-1,0)||
第一题译成中文是这样的:
椭圆是到两个定点的距离之和为常数的点的轨迹。请证明到定点(1,0)和(-1,0)距离之和为2根2的点所组成的椭圆可以表示为:(x^2)/(a^2) + (y^2)/(b^2) = 1,其中a和b是待定常数。(你的题目肯定打错了,按照那样的两个点写出来的椭圆方程是有交叉项的,所以我默认是焦点在x轴上)
由于题目是英文的,我就用英文解了,其中的sqrt()表示开平方。
Solution:Set a point on the ellipse P with cordinates (x,y)
Then,|PA|=sqrt[(x-1)^2+y^2], |PB|=sqrt[(x+1)^2+y^2], where |PA| and |PB| stand for the distances between P&A, and P&B.
According to the definition of a ellipse:|PA|+|PB|=2sqrt(2)
Thus,sqrt[(x-1)^2+y^2]+sqrt[(x+1)^2+y^2]=2sqrt(2)
sqrt(x^2+y^2+1-2x)+sqrt(x^2+y^2+1+2x)=2sqrt(2)
square both sides of this equation,
2x^2+2y^2+2+2sqrt[(x^2+y^2+1)^2-4x^2]=8
(x^2+y^2+1)^2-4x^2=[3-(x^2+y^2)]^2
(x^2+y^2)^2+2(x^2+y^2)+1-4x^2=(x^2+y^2)^2-6(x^2+y^2)+9
counteract the same parts of of the two sides and make some ajustments
we have:
4x^2+8y^2=8
thus, x^2/2+y^2=1, which meets the form of the equation given, where a=sqrt(2),b=1.
下面看第二题
a)