A) \[\frac{({{a}^{2}}-{{b}^{2}})ab}{{{a}^{2}}+{{b}^{2}}}\]
B) \[\frac{({{a}^{2}}-{{b}^{2}})}{({{a}^{2}}+{{b}^{2}})ab}\]
C) \[\frac{({{a}^{2}}-{{b}^{2}})}{ab({{a}^{2}}+{{b}^{2}})}\]
D) \[\frac{({{a}^{2}}+{{b}^{2}})}{({{a}^{2}}-{{b}^{2}})ab}\]
Correct Answer: C
Solution :
\[P\left( \frac{a}{\sqrt{2}},\,\frac{b}{\sqrt{2}} \right){{P}_{1}}=\frac{\sqrt{2}ab}{{{a}^{2}}+{{b}^{2}}}\,;\,\,{{p}_{2}}=\frac{{{a}^{2}}-{{b}^{2}}}{\sqrt{2}\,({{a}^{2}}+{{b}^{2}})}\]\[\Rightarrow \,\,{{p}_{1}}{{p}_{2}}=\] result \[T:\,\,\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1\] \[{{p}_{1}}=\left| \frac{ab}{\sqrt{{{b}^{2}}{{\cos }^{2}}\,\theta +{{a}^{2}}{{\sin }^{2}}\theta }} \right|\,\] ?(1) \[{{N}_{1}}=\frac{ax}{\cos \theta }\,-\frac{by}{\sin \theta }\,={{a}^{2}}-{{b}^{2}}\] \[{{p}_{2}}=\left| \frac{({{a}^{2}}-{{b}^{2}})\sin \theta \cos \theta }{\sqrt{{{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta }}\, \right|\] ?(2) \[{{p}_{1}}{{p}_{2}}=\frac{ab({{a}^{2}}-{{b}^{2}})}{2\left( \frac{{{a}^{2}}}{2}+\frac{{{b}^{2}}}{2} \right)}\]when \[\theta =\frac{\pi }{4};\,\,{{p}_{1}}{{p}_{2}}\,=\frac{ab({{a}^{2}}-{{b}^{2}})}{{{a}^{2}}+{{b}^{2}}}\]You need to login to perform this action.
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