A) \[y+x=\pm 1\]
B) \[y-x=\pm 1\]
C) \[3x+4y=\pm 1\]
D) \[3x-4y=\pm 1\]
Correct Answer: B
Solution :
The tangent at \[(h,k)\] is \[\frac{x}{4/h}-\frac{y}{3/k}=1\] \[\therefore \frac{4}{h}=\frac{3}{k}\]Þ\[\frac{h}{k}=\frac{4}{3}\] .....(i) and \[3{{h}^{2}}-4{{k}^{2}}=12\] ....(ii) As point \[(h,k)\]lies on it, using (i) and (ii), we get the tangent as \[y-x=\pm 1\].
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