A) \[y=x+1\]
B) \[y=x-1\]
C) \[y=x+2\]
D) \[y=x-2\]
Correct Answer: A
Solution :
\[\frac{{{x}^{2}}}{3}-\frac{{{y}^{2}}}{2}=1\] \[\because \] Equation of tangent are equally inclined to the axis i.e., \[\tan \theta =1=m\]. \ Eq. of tangent \[y=mx+\sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}\] Given eq. \[\frac{{{x}^{2}}}{3}-\frac{{{y}^{2}}}{2}=1\] is a eq. of hyperbola which is of form \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\]. Now, on comparing \[{{a}^{2}}=3\], \[{{b}^{2}}=2\] \ \[y=1.x+\sqrt{3\times {{(1)}^{2}}-2}\] Þ \[y=x+1\].
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