Conic Sections Question 449
Question: 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{PN^{2}}{AN\ .\ A’N} $ is
Options:
A) $ \frac{b^{2}}{a^{2}} $
B) $ \frac{a^{2}}{b^{2}} $
C) $ a^{2}+b^{2} $
D) 1
Show Answer
Answer:
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}}}{ANA’N}=\frac{b^{2}{{\sin }^{2}}\theta }{a^{2}(1-\cos \theta )(1+\cos \theta )}=\frac{b^{2}}{a^{2}} $ .
BETA
