A) \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}-{{b}^{2}}\]
B) \[{{x}^{2}}-{{y}^{2}}={{a}^{2}}-{{b}^{2}}\]
C) \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}\]
D) \[{{x}^{2}}-{{y}^{2}}={{a}^{2}}+{{b}^{2}}\]
Correct Answer: C
Solution :
Let point be \[(h,k)\]their pair of tangent will be \[\left( \frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}-1 \right)\,\left( \frac{{{h}^{2}}}{{{a}^{2}}}+\frac{{{k}^{2}}}{{{b}^{2}}}-1 \right)={{\left( \frac{hx}{{{a}^{2}}}+\frac{yk}{{{b}^{2}}}-1 \right)}^{2}}\] Pair of tangents will be perpendicular, if coefficient of \[{{x}^{2}}\]+ coefficient of \[{{y}^{2}}=0\] Þ\[\frac{{{k}^{2}}}{{{a}^{2}}{{b}^{2}}}+\frac{{{h}^{2}}}{{{a}^{2}}{{b}^{2}}}=\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}\]Þ \[{{h}^{2}}+{{k}^{2}}={{a}^{2}}+{{b}^{2}}\] Replace \[(h,k)\]by\[(x,y)\]Þ\[{{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}\].
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