The centre of an ellipse is C,PN is any ordin | vedclass

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(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Let the coordinates of point $P$ be $(a \cos 	heta, b \sin 	heta)$.Th

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$\frac{b^2}{a^2}$

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(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Let the coordinates of point $P$ be $(a \cos 	heta, b \sin 	heta)$.The coordinates of the end points of the major axis are $A \equiv (a, 0)$ and $A' \equiv (-a, 0)$.Since $PN$ is the ordinate of $P$,the coordinates of $N$ are $(a \cos 	heta, 0)$.Therefore,$PN = b \sin 	heta$.$AN = a - a \cos 	heta = a(1 - \cos 	heta)$.$A'N = a \cos 	heta - (-a) = a(1 + \cos 	heta)$.Now,calculate the ratio:$\frac{PN^2}{AN \cdot A'N} = \frac{(b \sin 	heta)^2}{a(1 - \cos 	heta) \cdot a(1 + \cos 	heta)}$$= \frac{b^2 \sin^2 	heta}{a^2(1 - \cos^2 	heta)}$$= \frac{b^2 \sin^2 	heta}{a^2 \sin^2 	heta}$$= \frac{b^2}{a^2}$.

The centre of an ellipse is $C$,$PN$ is any ordinate,and $A$,$A'$ are the end points of the major axis. Then the value of $\frac{PN^2}{AN \cdot A'N}$ is

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