A) \[p\cos \alpha +a=0\]
B) \[p\cos \alpha -a=0\]
C) \[a\cos \alpha +p=0\]
D) \[a\cos \alpha -p=0\]
Correct Answer: A
Solution :
\[x\cos \alpha +y\sin \alpha -p=0\] .?.(i) \[2ax-y{{y}_{1}}+2a({{x}_{1}}+2a)=0\] .?.(ii) From (i) and (ii), \[\frac{\cos \alpha }{2a}=\frac{\sin \alpha }{-y}=\frac{-p}{2a(x+2a)}\] Þ \[y=-2\tan \alpha \] and \[x=-p\sec \alpha -2a\] \[\therefore {{y}^{2}}=4a(x+a)\]Þ\[4{{a}^{2}}{{\tan }^{2}}\alpha =-4a(p\sec \alpha +a)\] Þ \[p\cos \alpha +a=0\].
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