If a chord joining two points whose eccentric angles are - - cut the major axis of the ellipse x 2 - a 2 - y 2 - b 2 -1- at a distance d from the center- then tan- 2 tan- 2 -

The correct option is D #160; d a / d+a Equation of chord joining P(α ) and Q(β ) is : x / a cos(α +β #160; / 2 )+ y / b si ...

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Standard XII

Physics

Relative Velocity

If a chord jo...

Question

If a chord joining two points whose eccentric angles are $α, β$ cut the major axis of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , at a distance $d$ from the center, then $tan \frac{α}{2} tan \frac{β}{2} =$

\frac{d + a}{d - a}

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\frac{d - a}{d + a}

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\frac{a - d}{a + d}

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None of these

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Solution

The correct option is D $\frac{d - a}{d + a}$
$E q u a t i o n o f c h o r d j o i n i n g P (α) a n d Q (β) i s : \frac{x}{a} c o s (\frac{α + β}{2}) + \frac{y}{b} s i n (\frac{α + β}{2}) = c o s (\frac{α - β}{2}) X - i n t e r c e p t d = a \frac{c o s (\frac{α - β}{2})}{c o s (\frac{α + β}{2})} O r, \frac{d}{a} = \frac{c o s (\frac{α - β}{2})}{c o s (\frac{α + β}{2})} B y c o m p o n e n d o a n d d i v i d e n d o w e g e t : \frac{d + a}{d - a} = \frac{c o s (\frac{α - β}{2}) + c o s (\frac{α + β}{2})}{c o s (\frac{α - β}{2}) - c o s (\frac{α + β}{2})} O r, \frac{d + a}{d - a} = \frac{2 c o s (\frac{α}{2}) * c o s (\frac{β}{2})}{2 s i n (\frac{α}{2}) * s i n (\frac{β}{2})} O r, t a n (\frac{α}{2}) t a n (\frac{β}{2}) = \frac{d - a}{d + a}$

Option [B]

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If a chord joining two points whose eccentric angles are α,β cut the major axis of the ellipse x2a2+y2b2=1, at a distance d from the center, then tanα2tanβ2=

If a chord joining two points whose eccentric angles are $α, β$ cut the major axis of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , at a distance $d$ from the center, then $tan \frac{α}{2} tan \frac{β}{2} =$