A) \[x-y+1=0\]
B) \[x-y+2=0\]
C) \[x+y-1=0\]
D) \[x+y-2=0\]
Correct Answer: A
Solution :
Let equation of tangent parallel to \[y=x\] drawn to\[\frac{{{x}^{2}}}{3}-\frac{{{y}^{2}}}{2}=1\]is\[y=x+c\] \[(\because \,\,m=1)\] We know that, for tangency \[{{c}^{2}}={{a}^{2}}{{m}^{2}}-{{b}^{2}}\] \[\Rightarrow \] \[{{c}^{2}}=3\cdot 1-2\] \[(a=3,\,\,b=2\]from given hyperbola) \[\Rightarrow \] \[{{c}^{2}}=1\] \[\Rightarrow \] \[c=\pm 1\] Hence, equation of tangent is \[y=x\pm 1\] \[\Rightarrow \] \[x-y=\pm 1\] \[\Rightarrow \] \[x-y\mp 1=0\]
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