求直线{x=1+4/5t(t为参数)被曲线ρ=√2cos(θ-π/4)所截的弦长 y=-1-3/5t

ρ=√2cos(θ-π/4)=cosθ-sinθ
ρ^2=ρcosθ-ρsinθ
x^2+y^2=x-y
将x=1+(4/5)t、y=-1-(3/5)t代入曲线方程得：t^2+7t/5=0
t1=-7/5、t2=0
所求弦长为：|t1-t2|=7/5.
.