If - 1 - 2 are the eccentric angles of the extremeties of a focal chard other than the vertices of the ellipse x2-a2 - y2-b2 - 1 a - b and e its eccentricity- Then show that e cos- 1 - - 2-2 - cos - 1 - - 2-2-

Let the co ordinates of the end points of the chord of the ellipse be P(acosθ_1,bsinθ_1) and #160; Q(acosθ_2,bsinθ_2) .Since this being the focal ch ...

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Byju's Answer

Standard XII

Mathematics

Parametric Representation: Ellipse

If θ 1 θ 2 ...

Question

If $θ_{1} θ_{2}$ are the eccentric angles of the extremeties of a focal chard (other than the vertices of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $(a > b)$ and $e$ its eccentricity. Then show that $e cos \frac{(θ_{1} + θ_{2})}{2} = cos \frac{θ_{1} - θ_{2}}{2}$ .

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Solution

Let the co-ordinates of the end points of the chord of the ellipse be $P (a cos θ_{1}, b sin θ_{1})$ and $Q (a cos θ_{2}, b sin θ_{2})$ .
Since this being the focal chord, with out loss of generality we assume that the chord $P Q$ passes through the focus $(a e, 0)$ .
Now the equation of the chord be
$\frac{y - b sin θ_{2}}{b (sin θ_{1} - sin θ_{2})} = \frac{x - a cos θ_{2}}{a (cos θ_{1} - cos θ_{2})}$
or, $\frac{y - b sin θ_{2}}{2 b (sin \frac{θ_{1} - θ_{2}}{2} . cos \frac{θ_{1} + θ_{2}}{2})} = \frac{x - a cos θ_{2}}{2 a (sin \frac{θ_{1} + θ_{2}}{2} sin \frac{θ_{2} - θ_{1}}{2})}$
or, $\frac{y - b sin θ_{2}}{b (cos \frac{θ_{1} + θ_{2}}{2})} = \frac{x - a cos θ_{2}}{- a (sin \frac{θ_{2} + θ_{1}}{2})}$ , this straight line passes through $(a e, 0)$ ,
then ,
$\frac{0 - b sin θ_{2}}{b (cos \frac{θ_{1} + θ_{2}}{2})} = \frac{a e - a cos θ_{2}}{- a (sin \frac{θ_{2} + θ_{1}}{2})}$ ,
or, $e cos \frac{θ_{1} + θ_{2}}{2} = cos θ_{2} cos \frac{θ_{1} + θ_{2}}{2} + sin θ_{2} . sin \frac{θ_{1} + θ_{2}}{2}$
or, $e cos \frac{θ_{1} + θ_{2}}{2} = cos \frac{θ_{1} - θ_{2}}{2}$

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If θ1θ2 are the eccentric angles of the extremeties of a focal chard (other than the vertices of the ellipse x2a2+y2b2=1 (a>b) and e its eccentricity. Then show that ecos(θ1+θ2)2=cosθ1−θ22.

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