A) \[{{a}^{2}}\]
B) \[{{b}^{2}}\]
C) \[{{a}^{2}}+{{b}^{2}}\]
D) \[{{a}^{2}}-{{b}^{2}}\]
Correct Answer: B
Solution :
\[\left( \frac{b\sqrt{{{a}^{2}}-{{b}^{2}}}\cos \theta +0-ab}{\sqrt{{{b}^{2}}{{\cos }^{2}}\theta +{{a}^{2}}{{\sin }^{2}}\theta }} \right)\,\left( \frac{-b\sqrt{{{a}^{2}}-{{b}^{2}}}\cos \theta -ab}{\sqrt{{{b}^{2}}{{\cos }^{2}}\theta +{{a}^{2}}{{\sin }^{2}}\theta }} \right)\] \[=\frac{-[{{b}^{2}}({{a}^{2}}-{{b}^{2}}){{\cos }^{2}}\theta -{{a}^{2}}{{b}^{2}}]}{({{b}^{2}}{{\cos }^{2}}\theta +{{a}^{2}}{{\sin }^{2}}\theta )}\] \[=\frac{{{b}^{2}}[{{a}^{2}}-{{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta ]}{{{b}^{2}}{{\cos }^{2}}\theta +{{a}^{2}}{{\sin }^{2}}\theta }\] \[=\frac{{{b}^{2}}[{{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta ]}{{{b}^{2}}{{\cos }^{2}}\theta +{{a}^{2}}{{\sin }^{2}}\theta }\]=\[{{b}^{2}}\]. Trick: Let \[a=2,b=1\]and\[\theta =\frac{\pi }{2}\], then the points are \[(\pm \sqrt{3},\,0)\] and the line is y = 1. Length from \[(\sqrt{3},0)\] on \[y=1\] is 1 and that of from \[(-\sqrt{3},0)\] is also 1. Hence product is\[1\times 1=1\], which is given by (b).
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