A) \[{{p}^{2}}={{a}^{2}}{{\sin }^{2}}\alpha +{{b}^{2}}{{\cos }^{2}}\alpha \]
B) \[{{p}^{2}}={{a}^{2}}+{{b}^{2}}\]
C) \[{{p}^{2}}={{b}^{2}}{{\sin }^{2}}\alpha +{{a}^{2}}{{\cos }^{2}}\alpha \]
D) None of these
Correct Answer: C
Solution :
\[y=-x\cot \alpha +\frac{p}{\sin \alpha }\]is tangent to \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1,\] if \[\frac{p}{\sin \alpha }=\pm \sqrt{{{b}^{2}}+{{a}^{2}}{{\cot }^{2}}\alpha }\] or \[{{p}^{2}}={{b}^{2}}{{\sin }^{2}}\alpha +{{a}^{2}}{{\cos }^{2}}\alpha \].
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