第16章 16.22

(1)

$x^2 = 4y$

$x^2 = 4py$

$p = 1$

焦点$(0,1)$ 準線$y = -1$

(2)

$y = x^2$

$x^2 = 4py$

$p = \dfrac{1}{4}$

焦点$\left(0, \dfrac{1}{4}\right)$ 準線 $y = -\dfrac{1}{4}$

(3)

$y = ax^2$

$x^2 = 4py$

$p = \dfrac{1}{4a}$

焦点$\left(0, \dfrac{1}{4a}\right)$ 準線 $y = -\dfrac{1}{4a}$

(4)

$y = a(x-p)^2 + q$ $y - q = a(x-p)^2$ これは$y = ax^2$を$(p,q)$平行移動した者なので

(3)から

焦点$\left(p, \dfrac{1}{4a} + q\right)$ 準線 $y = -\dfrac{1}{4a} + q$