第7章 7.21

$\sqrt{x} + \sqrt{y} = \sqrt{a}$

接点 $P(m, (\sqrt{a} - \sqrt{m})^2)$ とおくと

$\frac{dx}{2\sqrt{x}} + \frac{dy}{2\sqrt{y}} = 0 \qquad \frac{dy}{2\sqrt{y}} = -\frac{dx}{2\sqrt{x}} \qquad \frac{dy}{dx} = y' = -\frac{\sqrt{y}}{\sqrt{x}} = \frac{\sqrt{x} - \sqrt{a}}{\sqrt{x}}$

$x = m$ のとき $y' = \dfrac{\sqrt{m} - \sqrt{a}}{\sqrt{m}}$

接線：$y - (\sqrt{a} - \sqrt{m})^2 = \dfrac{\sqrt{m} - \sqrt{a}}{\sqrt{m}}(x - m) \qquad y = \dfrac{\sqrt{m} - \sqrt{a}}{\sqrt{m}} x + a - \sqrt{am}$

x軸との交点 $A\left(\dfrac{m\sqrt{a} - a\sqrt{m}}{\sqrt{m} - \sqrt{a}}, 0\right)$ y軸との交点 $B(0, a - \sqrt{am})$

$OA + OB = \frac{m\sqrt{a} - a\sqrt{m}}{\sqrt{m} - \sqrt{a}} + a - \sqrt{am} = \frac{m\sqrt{a} - a\sqrt{m} + (a - \sqrt{am})(\sqrt{m} - \sqrt{a})}{\sqrt{m} - \sqrt{a}}$

$= \frac{m\sqrt{a} - a\sqrt{m} + a\sqrt{m} - a\sqrt{a} - m\sqrt{a} + a\sqrt{m}}{\sqrt{m} - \sqrt{a}} = \frac{a\sqrt{m} - a\sqrt{a}}{\sqrt{m} - \sqrt{a}} = a$

よって一定である