A) \[4{{x}^{2}}{{y}^{2}}+2({{x}^{2}}-{{a}^{2}})({{y}^{2}}-{{b}^{2}})=k{{({{x}^{2}}-{{a}^{2}})}^{2}}\]
B) \[4{{x}^{2}}{{y}^{2}}-2({{x}^{2}}-{{a}^{2}})({{y}^{2}}-{{b}^{2}})=k{{({{x}^{2}}-{{a}^{2}})}^{2}}\]
C) \[4{{x}^{2}}{{y}^{2}}-2({{x}^{2}}-{{a}^{2}})({{y}^{2}}-{{b}^{2}})=k{{({{x}^{2}}+{{a}^{2}})}^{2}}\]
D) None of these
Correct Answer: B
Solution :
[b] If the tangent \[y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}\] passes through \[({{x}_{1}},\,\,{{y}_{1}})\] then \[{{({{y}_{1}}-m{{x}_{1}})}^{2}}={{a}^{2}}{{m}^{2}}+{{b}^{2}}\] or \[({{x}^{2}}_{1}-{{a}^{2}}){{m}^{2}}-2{{x}_{1}}{{y}_{1}}m+{{y}^{2}}_{1}-{{b}^{2}}=0.\] If The roots be \[{{m}_{1}}\] and \[{{m}_{2}}\] then \[{{m}_{1}}+{{m}_{2}}=\frac{2{{x}_{1}}{{y}_{1}}}{{{x}^{2}}_{1}-{{a}^{2}}}\] and \[{{m}_{1}}{{m}_{2}}=\frac{{{y}^{2}}_{1}-{{b}^{2}}}{{{x}^{2}}_{1}-{{a}^{2}}}\] Given: \[{{\tan }^{2}}{{\theta }_{1}}+{{\tan }^{2}}{{\theta }_{2}}=\] constant \[=\,\,k\] (say) \[\Rightarrow {{m}^{2}}_{1}+{{m}^{2}}_{2}=k\] or \[{{({{m}_{1}}+{{m}_{2}})}^{2}}-2{{m}_{1}}{{m}_{2}}=k\] \[\Rightarrow \frac{4{{x}^{2}}{{y}^{2}}}{{{({{x}^{2}}_{1}-{{a}^{2}})}^{2}}}-\frac{2({{y}^{2}}_{1}-{{b}^{2}})}{({{x}^{2}}_{1}-{{a}^{2}})}=k\] or \[4{{x}^{2}}_{1}{{y}^{2}}_{1}-2({{x}^{2}}_{1}-{{a}^{2}})({{y}^{2}}_{1}-{{b}^{2}})=k{{({{x}^{2}}_{1}-{{a}^{2}})}^{2}}\] Hence locus of P is \[4{{x}^{2}}{{y}^{2}}-2({{x}^{2}}-{{a}^{2}})({{y}^{2}}-{{b}^{2}})=k{{({{x}^{2}}-{{a}^{2}})}^{2}}\]
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