A) \[x-y-1=0\]
B) \[x-y+2=0\]
C) \[x+y-1=0\]
D) \[x+y+2=0\]
Correct Answer: A
Solution :
Given hyperbola is, \[\frac{{{x}^{2}}}{3}-\frac{{{y}^{2}}}{2}=1\] .....(i) Equation of tangent parallel to \[y-x+5=0\] is \[y-x+\lambda =0\] \[\Rightarrow \]\[y=x-\lambda \] .....(ii) If line (ii) is a tangent to hyperbola (i), then \[-\lambda =\pm \sqrt{3\times 1-2}\] (from \[c=\pm \sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}\]) \[-\lambda =\pm \,1\Rightarrow \lambda =-1,\,+1\]. Put the values of \[\lambda \] in (ii), we get \[x-y-1=0\] and \[x-y+1=0\] are the required tangents.You need to login to perform this action.
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