8.
設\( a>0 \)，\( O(0,0) \)為原點。在拋物線\( ay=a^2-x^2 \)上取一點\( P(s,t) \)，\( s>0 \)。過\( P \)點作拋物線之切線交\( x \)軸，\( y \)軸於\( Q、R \)兩點。當\( P \)點變動時，求\( \Delta OQR \)面積的最小值。
[解答]
\({x^2} = - a(y - a)\)
令\(P(at,a - a{t^2})\)
\( \displaystyle y' = - \frac{2}{a}x\)，故過P點之切線斜率為\( - 2t\)
此切線方程式為\( \displaystyle \frac{x}{{\displaystyle \frac{{a + a{t^2}}}{{2t}}}} + \frac{y}{{a + a{t^2}}} = 1\)
\( \displaystyle \Delta = \frac{1}{2}\frac{{a + a{t^2}}}{{2t}}(a + a{t^2}) = \frac{1}{4}\frac{{{{(a + a{t^2})}^2}}}{t}\)
\( \displaystyle \Delta ' = \frac{1}{4}\frac{{2(a + a{t^2}) \cdot 2at \cdot t - (a + a{t^2})2}}{{{t^2}}} = 0\)
解得\( \displaystyle t = \frac{1}{{\sqrt 3 }}\)代入
\( \displaystyle \Delta = \frac{1}{4}\sqrt 3 \cdot {a^2}{(1 + \frac{1}{3})^2} = \frac{{4\sqrt 3 }}{9}{a^2}\)
