A) (4, 3)
B) (3, 4)
C) (4, ?3)
D) None of these
Correct Answer: A
Solution :
The equation of the line and hyperbola are \[y=x-1\] .....(i) \[3{{x}^{2}}-4{{y}^{2}}=12\] .....(ii) From (i) and (ii), we get \[3{{x}^{2}}-4{{(x-1)}^{2}}=12\] \[\Rightarrow 3{{x}^{2}}-4({{x}^{2}}-2x+1)=12\] or \[{{x}^{2}}-8x+16=0\]Þ \[x=4\] From (i), \[y=3\]. So point of contact is\[(4,3)\].You need to login to perform this action.
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