Question

文科数学考坐标系与参数方程吗

Answer 1

文科数学考坐标系与参数方程，都考。
必考啊
虽然是选作

Answer 2

椭圆:x²/a²+y²/b²=1
若oa,ob有一个斜率不存在,则s△aob=(1/2)a×b=ab/2
若oa,ob斜率都存在,
设oa:y=kx,
ob:y=-x/k
y=kx和x²/a²+y²/b²=1联立,
(1/a²+k²/b²)x²=1,
xa²=1/(1/a²+k²/b²)
同理,
xb²=1/(1/a²+1/k²b²)
s△aob=(1/2)oa×ob=(1/2)×√(1+k²)×|xa|×√(1+(-1/k)²)×|xb|
=(1/2)×(1+k²)/√k²×√1/(1/a²+k²/b²)×√1/(1/a²+1/k²b²)
=(1/2)ab×√[(k^4+2k²+1)/(k^4+a²k²/b²+b²k²/a²+1)]
=(1/2)ab×√[1-(a²k²/b²+b²k²/a²-2k²)/(k^4+a²k²/b²+b²k²/a²+1)]
=(1/2)ab×√[1-(a²/b²+b²/a²-2)/(k²+a²/b²+b²/a²+1/k²)]
∵k²+1/k²>=2
∴0<(a²/b²+b²/a²-2)/(k²+a²/b²+b²/a²+1/k²)<=(a²/b²+b²/a²-2)/(a²/b²+b²/a²+2)
∴4/(a²/b²+b²/a²+2)<=[1-(a²/b²+b²/a²-2)/(k²+a²/b²+b²/a²+1/k²)]<1
即[2/(a/b+b/a)]²<=[1-(a²/b²+b²/a²-2)/(k²+a²/b²+b²/a²+1/k²)]<1
∴ab/(a/b+b/a)<=s△aob<ab/2
即a²b²/(a²+b²)<=s△aob<ab/2
综上,s△aob最大值为ab/2,
最小值为a²b²/(a²+b²)