A) \[\frac{{{b}^{2}}}{{{a}^{2}}}\]
B) \[\frac{{{a}^{2}}}{{{b}^{2}}}\]
C) \[{{a}^{2}}+{{b}^{2}}\]
D) 1
Correct Answer: A
Solution :
Let ellipse be \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] \[P=(a\cos \theta ,\,b\sin \theta ),\,A\,\text{ and}\,A'\equiv (\pm a,0),\,\,N\equiv (a\cos \theta ,0),\] \[PN=b\sin \theta ,\]\[AN=a(1-\cos \theta ),\] \[A'N=a(1+\cos \theta )\] \[\frac{{{(PN)}^{2}}}{AN\,A'N}=\frac{{{b}^{2}}{{\sin }^{2}}\theta }{{{a}^{2}}(1-\cos \theta )(1+\cos \theta )}=\frac{{{b}^{2}}}{{{a}^{2}}}\].You need to login to perform this action.
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