question_answer

The centre of an ellipse is C and PN is any ordinate and A, A? are the end points of major axis, then the value of \[\frac{P{{N}^{2}}}{AN\ .\ A'N}\] is

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 \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] \[P=(a\cos \theta ,\,b\sin \theta ),\,A\,\text{ and}\,A'\equiv (\pm a,0),\,\,N\equiv (a\cos \theta ,0),\]                   \[PN=b\sin \theta ,\]\[AN=a(1-\cos \theta ),\]                   \[A'N=a(1+\cos \theta )\]                    \[\frac{{{(PN)}^{2}}}{AN\,A'N}=\frac{{{b}^{2}}{{\sin }^{2}}\theta }{{{a}^{2}}(1-\cos \theta )(1+\cos \theta )}=\frac{{{b}^{2}}}{{{a}^{2}}}\].