A) \[{{a}^{2}}{{\cos }^{2}}\alpha +{{b}^{2}}{{\sin }^{2}}\alpha ={{p}^{2}}\]
B) \[{{a}^{2}}{{\cos }^{2}}\alpha -{{b}^{2}}{{\sin }^{2}}\alpha ={{p}^{2}}\]
C) \[{{a}^{2}}{{\sin }^{2}}\alpha +{{b}^{2}}{{\cos }^{2}}\alpha ={{p}^{2}}\]
D) \[{{a}^{2}}{{\sin }^{2}}\alpha -{{b}^{2}}{{\cos }^{2}}\alpha ={{p}^{2}}\]
Correct Answer: B
Solution :
\[x\cos \alpha +y\sin \alpha =p\Rightarrow y=-\cot \alpha .\,\,x+p\text{cosec }\alpha \] It is tangent to the hyperbola \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] Therefore, \[{{p}^{2}}\text{cose}{{\text{c}}^{2}}\alpha ={{a}^{2}}{{\cot }^{2}}\alpha -{{b}^{2}}\] \[\Rightarrow {{a}^{2}}{{\cos }^{2}}\alpha -{{b}^{2}}{{\sin }^{2}}\alpha ={{p}^{2}}\].
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