A) \[{{y}^{2}}+{{b}^{2}}={{c}^{2}}({{x}^{2}}-{{a}^{2}})\]
B) \[{{y}^{2}}+{{b}^{2}}={{c}^{2}}({{x}^{2}}+{{a}^{2}})\]
C) \[a{{x}^{2}}+b{{y}^{2}}={{c}^{2}}\]
D) None of these
Correct Answer: A
Solution :
Let \[(h,k)\]be the point of intersection. By\[S{{S}_{1}}={{T}^{2}}\], \[\left( \frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}-1 \right)\text{ }\left( \frac{{{h}^{2}}}{{{a}^{2}}}-\frac{{{k}^{2}}}{{{b}^{2}}}-1 \right)={{\left[ \frac{hx}{{{a}^{2}}}-\frac{ky}{{{b}^{2}}}-1 \right]}^{2}}\] Þ\[{{x}^{2}}\left[ \frac{{{h}^{2}}}{{{a}^{4}}}-\frac{{{k}^{2}}}{{{a}^{2}}{{b}^{2}}}-\frac{1}{{{a}^{2}}}-\frac{{{h}^{2}}}{{{a}^{4}}} \right]-{{y}^{2}}\left[ \frac{{{h}^{2}}}{{{a}^{2}}{{b}^{2}}}-\frac{{{k}^{2}}}{{{b}^{4}}}-\frac{1}{{{b}^{2}}}+\frac{{{k}^{2}}}{{{b}^{4}}} \right]+...=0\] We know that, \[{{m}_{1}}{{m}_{2}}=\frac{\text{Coefficent}\,\,\text{of}\,\,{{x}^{2}}}{\text{Coefficent}\,\,\text{of}\,\,{{y}^{2}}}\] \[\Rightarrow \]\[{{m}_{1}}{{m}_{2}}=\frac{\frac{{{k}^{2}}}{{{a}^{2}}{{b}^{2}}}+\frac{1}{{{a}^{2}}}}{\frac{{{h}^{2}}}{{{a}^{2}}{{b}^{2}}}-\frac{1}{{{b}^{2}}}}={{c}^{2}}\] \[\Rightarrow \]\[\left( \frac{{{k}^{2}}+{{b}^{2}}}{{{h}^{2}}-{{a}^{2}}} \right)={{c}^{2}}\] or \[({{y}^{2}}+{{b}^{2}})={{c}^{2}}({{x}^{2}}-{{a}^{2}})\].
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