回復 1# Superconan 的帖子
填充第3題
設橢圓:\(\displaystyle \frac{x^2}{5^2}+\frac{y^2}{4^2}=1\),點\(P(x_0,y_0)\)為橢圓上第一象限中的一點,過點\(P\)作切線交\(x\)軸交於點\(A\),交\(y\)軸於點\(B\),當\(\overline{AB}\)有最小值時的切線方程式為\(y=ax+b\),則數對\((a,b)=\) 。
\(P\left( 5\cos \theta ,4\sin \theta \right)\)
直線\(AB\)之方程式為\(\frac{\cos \theta }{5}x+\frac{\sin \theta }{4}y=1\)
\(\begin{align}
& A\left( \frac{5}{\cos \theta },0 \right),B\left( 0,\frac{4}{\sin \theta } \right) \\
& \overline{AB}=\sqrt{{{\left( \frac{5}{\cos \theta } \right)}^{2}}+{{\left( \frac{4}{\sin \theta } \right)}^{2}}} \\
& \left[ {{\left( \frac{5}{\cos \theta } \right)}^{2}}+{{\left( \frac{4}{\sin \theta } \right)}^{2}} \right]\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)\ge {{\left( 5+4 \right)}^{2}}=81 \\
\end{align}\)
等號成立於\(\cos \theta =\frac{\sqrt{5}}{3},\sin \theta =\frac{2}{3}\)
直線\(AB\)之方程式為\(y=-\frac{2\sqrt{5}}{5}x+6\)