A) \[x\pm y=a\sqrt{2}\]
B) \[x\pm y=\pm a\sqrt{2}\]
C) \[x\pm y=2a\]
D) \[x+y=\pm 2a\]
Correct Answer: B
Solution :
Let the tangent be of form \[\frac{x}{{{x}_{1}}}+\frac{y}{{{y}_{1}}}=1\] and area of \[\Delta \] formed by it with coordinate axes is \[\frac{1}{2}{{x}_{1}}{{y}_{1}}={{a}^{2}}\] ?.(i) Again, \[{{y}_{1}}x+{{x}_{1}}y-{{x}_{1}}{{y}_{1}}=0\] Applying conditions of tangency \[\left| \frac{-{{x}_{1}}{{y}_{1}}}{\sqrt{x_{1}^{2}+y_{1}^{2}}} \right|\ =a\] or \[(x_{1}^{2}+y_{1}^{2})=\frac{x_{1}^{2}y_{1}^{2}}{{{a}^{2}}}\] ?.(ii) From (i) and (ii), we get\[{{x}_{1}},{{y}_{1}}\]; which gives equation of tangent as\[x\pm y=\pm a\sqrt{2}\]. Trick: There may be 4 tangents (as in figure). As the lines \[x\pm y=\pm a\sqrt{2}\] make triangle of area ?a? in all four quadrants.You need to login to perform this action.
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