If αand β re eccentric angles of the ends of a focal chord of the ellipse x2a2+y2b2=1 then tan⁡α2tan⁡β2 is equal to

The coordinates of the end points of the focal chord are (acos⁡α,bsin⁡α) and (acos⁡β,bsin⁡β) Therefore, theequation of the focal chord is xacos⁡α+β2+ybsin⁡α+β2=cos⁡α−β2 This passes through (ae,0). ∴ aeacos⁡α+β2=cos⁡α−β2⇒ cos⁡α+β2cos⁡α−β2=1e⇒ cos⁡α+β2−cos⁡α−β2cos⁡α+β2+cos⁡α−β2=1−e1+e⇒ −2sin⁡α2sin⁡β22cos⁡α2cos⁡β2=1−e1+e⇒ −tan⁡α2tan⁡β2=1−e1+e⇒tan⁡α2tan⁡β2=e−1e+1

If αand β re eccentric angles of the ends of a focal chord of the ellipse x2a2+y2b2=1 then tan⁡α2tan⁡β2 is equal to

The coordinates of the end points of the focal chord are (acos⁡α,bsin⁡α) and (acos⁡β,bsin⁡β) Therefore, theequation of the focal chord is xacos⁡α+β2+ybsin⁡α+β2=cos⁡α−β2 This passes through (ae,0). ∴ aeacos⁡α+β2=cos⁡α−β2⇒ cos⁡α+β2cos⁡α−β2=1e⇒ cos⁡α+β2−cos⁡α−β2cos⁡α+β2+cos⁡α−β2=1−e1+e⇒ −2sin⁡α2sin⁡β22cos⁡α2cos⁡β2=1−e1+e⇒ −tan⁡α2tan⁡β2=1−e1+e⇒tan⁡α2tan⁡β2=e−1e+1

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If $α$ and $β$ re eccentric angles of the ends of a focal chord of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ then $\tan \frac{α}{2} \tan \frac{β}{2}$ is equal to

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$\frac{1 - e}{1 + e}$

none of these

$\frac{e - 1}{e + 1}$

$\frac{e + 1}{e - 1}$

answer is B.

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Detailed Solution

The coordinates of the end points of the focal

chord are $(a \cos α, b \sin α)$ and $(a \cos β, b \sin β)$ Therefore, the

equation of the focal chord is

$\frac{x}{a} \cos (\frac{α + β}{2}) + \frac{y}{b} \sin (\frac{α + β}{2}) = \cos (\frac{α - β}{2})$

This passes through $(a e, 0)$ .

$\begin{array}{l} ∴ \frac{a e}{a} \cos (\frac{α + β}{2}) = \cos (\frac{α - β}{2}) \\ \Rightarrow \frac{\cos (\frac{α + β}{2})}{\cos (\frac{α - β}{2})} = \frac{1}{e} \\ \Rightarrow \frac{\cos (\frac{α + β}{2}) - \cos (\frac{α - β}{2})}{\cos (\frac{α + β}{2}) + \cos (\frac{α - β}{2})} = \frac{1 - e}{1 + e} \\ \Rightarrow - \frac{2 \sin \frac{α}{2} \sin \frac{β}{2}}{2 \cos \frac{α}{2} \cos \frac{β}{2}} = \frac{1 - e}{1 + e} \\ \Rightarrow - \tan \frac{α}{2} \tan \frac{β}{2} = \frac{1 - e}{1 + e} \Rightarrow \tan \frac{α}{2} \tan \frac{β}{2} = \frac{e - 1}{e + 1} \end{array}$