第1章 1.7

(1)

x=t+1x = t + 1 ①

y=3−2ty = 3 - 2t ②

①から

t=x−1t = x - 1 ①’

①’を②に代入して

y=3−2(x−1)=−2x+5y = 3 - 2(x-1) = -2x + 5 (直線)

(2)

x=t2+1x = t^2 + 1 ①

y=t+1y = t + 1 ②

②から

t=y−1t = y - 1 ②’

②’を①に代入して

x=(y−1)2+1x = (y-1)^2 + 1 (放物線)

(3)

x=t+1tx = t + \dfrac{1}{t} ①

y=t−1ty = t - \dfrac{1}{t} ②

①+②から

x+y=2t∴t=12(x+y)x + y = 2t \qquad \therefore t = \dfrac{1}{2}(x+y) ③

③を①に代入して

x=12(x+y)+112(x+y)=(x+y)2+42(x+y)x = \dfrac{1}{2}(x+y) + \dfrac{1}{\dfrac{1}{2}(x+y)} = \dfrac{(x+y)^2 + 4}{2(x+y)}

2x(x+y)=(x+y)2+4∴x2−y2=42x(x+y) = (x+y)^2 + 4 \qquad \therefore x^2 - y^2 = 4 (双曲線)

(4)

x=2+cos⁡t①x = 2 + \cos t \qquad ①

y=sin⁡t−1②y = \sin t - 1 \qquad ②

①から

x−2=cos⁡t①′x - 2 = \cos t \qquad ①'

②から

y+1=sin⁡t②′y + 1 = \sin t \qquad ②'

①’2^2+②’2^2から

(x−2)2+(y+1)2=cos⁡2t+sin⁡2t(x-2)^2 + (y+1)^2 = \cos^2 t + \sin^2 t

(x−2)2+(y+1)2=1(円)(x-2)^2 + (y+1)^2 = 1 \qquad (\text{円})

解説: r31bn1z