已知点$F$是椭圆$\dfrac{x^2}{25}+\dfrac{y^2}9=1$的左焦点，直线$AB$经过$F$且与椭圆交于$A,B$两点．若$O$为坐标原点，$\triangle AOB$的面积是$\dfrac 92$，求直线$AB$的斜率$k$．
解法一 设$AB:x=my-4$，则$$\dfrac{1}{25}(my-4)^2+\dfrac 19y^2=1,$$从而$$\left(\dfrac{m^2}{25}+\dfrac 19\right)y^2-\dfrac{8m}{25}y-\dfrac{9}{25}=0,$$从而$$S_{\triangle AOB}=\dfrac 12\cdot 4\cdot \dfrac{1}{\dfrac{m^2}{25}+\dfrac 19}\cdot \sqrt{\left(\dfrac{8m}{25}\right)^2-4\left(\dfrac{m^2}{25}+\dfrac 19\right)\cdot \left(-\dfrac{9}{25}\right)}=\dfrac 92,$$整理得$$81m^4-1150m^2-975=0,$$于是$$(m^2-15)(81m^2+65)=0,$$因此直线$AB$的斜率$k=\pm\dfrac{\sqrt{15}}{15}$．
解法二 设$\angle BFO=\theta$，则$$S_{\triangle AOB}=\dfrac 12 \cdot 4\cdot \left(\dfrac{9}{5-4\cos\theta}+\dfrac{9}{5+4\cos\theta}\right)\cdot \sin\theta =\dfrac 92,$$整理得$$40\sin\theta =25-16\cos^2\theta,$$即$$(4\sin\theta -1)(4\sin\theta-9)=0,$$因此$\sin\theta=\dfrac 14$，从而直线$AB$的斜率$k=\pm\dfrac{\sqrt{15}}{15}$．
解法三 利用仿射变换$$x'=x,y'=\dfrac 53y,$$将椭圆变成圆$x'^2+y'^2=25$，则$$S_{\triangle A'OB'}=\dfrac 92\cdot \dfrac 53=\dfrac{15}2.$$
设$O$到直线$A'B'$的距离为$OH$，则$$\dfrac 12\cdot 2\sqrt{25-OH^2}\cdot OH=\dfrac{15}2,$$即$$OH^2(25-OH^2)=\dfrac{225}{4},$$于是$OH^2=\dfrac 52$，从而$$\sin\angle B'F'O=\sqrt{\dfrac{5}{32}},$$从而$$\tan\angle B'F'O=\pm \sqrt{\dfrac{5}{27}},$$因此所求直线的斜率$k=\pm\dfrac{\sqrt{15}}{15}$．