A) \[{{m}_{1}}+{{m}_{2}}=\frac{24}{11}\]
B) \[{{m}_{1}}{{m}_{2}}=\frac{20}{11}\]
C) \[{{m}_{1}}+{{m}_{2}}=\frac{48}{11}\]
D) \[{{m}_{1}}{{m}_{2}}=\frac{11}{20}\]
Correct Answer: B
Solution :
The line through (6,2) is \[y-2=m(x-6)\]Þ\[y=mx+2-6m\] Now from condition of tangency, \[{{(2-6m)}^{2}}=25{{m}^{2}}-16\] Þ\[36{{m}^{2}}+4-24m-25{{m}^{2}}+16=0\] Þ\[11{{m}^{2}}-24m+20=0\] Obviously its roots are \[{{m}_{1}}\]and\[{{m}_{2}}\], therefore \[{{m}_{1}}+{{m}_{2}}=\frac{24}{11}\]and \[{{m}_{1}}{{m}_{2}}=\frac{20}{11}\].
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