If\[x\,\,\sin \theta -y\cos \theta =\sqrt{{{x}^{2}}+{{y}^{2}}}\]and\[\frac{{{\cos }^{2}}\theta }{{{a}^{2}}}+\frac{{{\sin }^{2}}\theta }{{{b}^{2}}}=\frac{1}{{{x}^{2}}+{{y}^{2}}},\]then the correct relation is [SSC (10+2) 2013] |
A) \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]
B) \[\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1\]
C) \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\]
D) \[\frac{{{x}^{2}}}{{{b}^{2}}}-\frac{{{y}^{2}}}{{{a}^{2}}}=1\]
Correct Answer: B
Solution :
Given, \[x\sin \theta -y\cos \theta =\sqrt{{{x}^{2}}+{{y}^{2}}}\] |
On squaring both sides, we get |
\[{{(x\sin \theta -y\cos \theta )}^{2}}={{(\sqrt{{{x}^{2}}+{{y}^{2}}})}^{2}}\] |
\[\Rightarrow \]\[{{x}^{2}}{{\sin }^{2}}\theta +{{y}^{2}}{{\cos }^{2}}\theta -2xy\] |
\[\sin \theta \cos \theta ={{x}^{2}}+{{y}^{2}}\] |
\[\Rightarrow \]\[{{x}^{2}}{{\sin }^{2}}\theta +{{y}^{2}}{{\cos }^{2}}\theta \] |
\[-\,\,2xy\sin \theta \cos \theta -{{x}^{2}}-{{y}^{2}}=0\] |
\[\Rightarrow \]\[{{x}^{2}}{{\sin }^{2}}\theta -{{x}^{2}}+{{y}^{2}}{{\cos }^{2}}\theta -{{y}^{2}}\] |
\[-\,\,2xy\sin \theta \cos \theta =0\] |
\[\Rightarrow \]\[{{x}^{2}}\,\,({{\sin }^{2}}\theta -1)+{{y}^{2}}\,\,({{\cos }^{2}}\theta -1)\] |
\[-\,\,2xy\sin \theta \cos \theta =0\] |
\[\Rightarrow \]\[-{{x}^{2}}{{\cos }^{2}}\theta -{{y}^{2}}{{\sin }^{2}}\theta \] |
\[-\,\,2xy\sin \theta \cos \theta =0\] |
\[\Rightarrow \]\[{{(x\cos \theta +y\,\,\sin \theta )}^{2}}=0\] \[[\because {{a}^{2}}+{{b}^{2}}+2ab={{(a+b)}^{2}}]\] |
\[\Rightarrow \] \[x\cos \theta +y\sin \theta =0\] |
\[\Rightarrow \] \[x\cos \theta =-\,y\sin \theta \] |
\[\Rightarrow \] \[\tan \theta =-\frac{x}{y}\] |
In \[\Delta ABC,\]\[\sin \theta =\frac{BC}{AC}\] |
\[\Rightarrow \]\[\sin \theta =\frac{x}{\sqrt{{{x}^{2}}+{{y}^{2}}}}\]or \[-\frac{x}{\sqrt{{{x}^{2}}+{{y}^{2}}}}\] |
Similarly, \[\cos \theta =-\frac{y}{\sqrt{{{x}^{2}}+{{y}^{2}}}}\]or \[\frac{y}{\sqrt{{{x}^{2}}+{{y}^{2}}}}\] |
Now, \[\frac{{{\cos }^{2}}\theta }{{{a}^{2}}}+\frac{{{\sin }^{2}}\theta }{{{b}^{2}}}=\frac{1}{{{x}^{2}}+{{y}^{2}}}\] |
\[\Rightarrow \]\[\frac{{{y}^{2}}}{({{x}^{2}}+{{y}^{2}})\cdot {{a}^{2}}}+\frac{{{x}^{2}}}{({{x}^{2}}+{{y}^{2}})\cdot {{b}^{2}}}=\frac{1}{{{x}^{2}}+{{y}^{2}}}\] |
\[\Rightarrow \]\[\frac{{{y}^{2}}}{{{a}^{2}}}+\frac{{{x}^{2}}}{{{b}^{2}}}=1\]\[\Rightarrow \]\[\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1\] |
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