英文数学题一道+函数一道1.An ellipse is the locations of points where the sum of the distances to two fixed points is constant.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)

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英文数学题一道+函数一道1.Anellipseisthelocationsofpointswherethesumofthedistancestotwofixedpointsisconstant.Sho

英文数学题一道+函数一道1.An ellipse is the locations of points where the sum of the distances to two fixed points is constant.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)
英文数学题一道+函数一道
1.An ellipse is the locations of points where the sum of the distances to two fixed points is constant.
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)||

英文数学题一道+函数一道1.An ellipse is the locations of points where the sum of the distances to two fixed points is constant.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)
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)
令y=0,则2f(x)=2x^2
所以f(x)=x^2
b)
令x=0,g(t)-g(-t)=0
所以g(x)是一个偶函数
令t=x,则g(2x)-g(0)=(2x)^2
由于g(0)的不确定性,所以g(x)=x^2+C,其中C是任意常数

1、我相题目应该是
…………………………………………………………………………
Show that the ellipse where the sum of the distances to (1,0) and (-1,0) is 2√2 can be written as
x2/a2 + y2/b2 = 1
where a and b are constants...

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1、我相题目应该是
…………………………………………………………………………
Show that the ellipse where the sum of the distances to (1,0) and (-1,0) is 2√2 can be written as
x2/a2 + y2/b2 = 1
where a and b are constants to be determined.
易得b=1,a=√2
2、
a)取y=0得2f(x)=2x2,所以f(x)=x2
b)取t=x得g(2x)-g(0)=4x2=(2x)2,设g(0)=c,则g(x)=x2+c

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