A) \[2x+3y=36\]
B) \[2x+3y+36=0\]
C) \[3x+2y=36\]
D) \[3x+2y+36=0\]
Correct Answer: B
Solution :
\[{{S}_{1}}\equiv {{x}^{2}}-108y=0\] \[T\equiv x{{x}_{1}}-2a(y+{{y}_{1}})=0\]Þ \[x{{x}_{1}}-54\left( y+\frac{x_{1}^{2}}{108} \right)=0\] \[{{S}_{2}}\equiv {{y}^{2}}-32x=0\] \[T\equiv y{{y}_{2}}-2a(x+{{x}_{2}})=0\] Þ \[y{{y}_{2}}-16\left( x+\frac{y_{2}^{2}}{32} \right)=0\] \\[\frac{{{x}_{1}}}{16}=\frac{54}{{{y}_{2}}}=\frac{-x_{1}^{2}}{y_{2}^{2}}=r\] Þ \[{{x}_{1}}=16r\] and \[{{y}_{2}}=\frac{54}{r}\] \\[\frac{-{{(16r)}^{2}}}{{{(54/r)}^{2}}}=r\] Þ \[r=-\frac{9}{4}\] \[{{x}_{1}}=-36,\ {{y}_{2}}=-24,\ {{y}_{1}}=\frac{{{(36)}^{2}}}{108}=12,\ {{x}_{2}}=18\]. \ Equation of common tangent \[(y-12)=\frac{-36}{54}(x+36)\Rightarrow 2x+3y+36=0\] Aliter: Using direct formula of common tangent\[y{{b}^{1/3}}+x\,{{a}^{1/3}}+{{(ab)}^{2/3}}=0\], where \[a=8\] and \[b=27\]. Hence the required tangent is\[3y+2x+36=0\].
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