If a, b are the eccentric angles of the
extremities of a focal chord of the ellipse
$x^{2} / 16 + y^{2} / 9 = 1$ then $\tan (α / 2) \tan (β / 2) =$

Question

If a, b are the eccentric angles of the extremities of a focal chord of the ellipse $$x2/16+y2/9=1$$ then  $$tan$$⁡$$($$α$$/2)$$$$tan$$⁡$$($$β$$/2)=$$

Infinity Learn · Accepted Answer

The eccentricity  $$e=1$$−$$916=74Let$$ $$P(4cos$$⁡α,$$3sin$$⁡α$$)$$  and $$Q(4cos$$⁡β,$$3sin$$⁡β$$)be$$ a focal chord of the ellipse passing through the focus at  (7$$,$$0) Then   $$3sin$$⁡β$$4cos$$⁡β−$$7=3sin$$⁡α$$4cos$$⁡α−$$7$$⇒     $$sin$$⁡$$($$α−β$$)$$$$sin$$⁡α−$$sin$$⁡β$$=74$$⇒     $$cos$$⁡[$$($$α−β$$)/2$$]$$cos$$⁡[$$($$α$$+$$β$$)/2$$]$$=74$$⇒     $$tan$$⁡α$$2tan$$⁡β$$2=7$$−$$47+4=23$$−$$87$$−$$9$$

If a, b are the eccentric angles of the
extremities of a focal chord of the ellipse
$x^{2} / 16 + y^{2} / 9 = 1$ then $\tan (α / 2) \tan (β / 2) =$

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If a, b are the eccentric angles of the extremities of a focal chord of the ellipse x2/16+y2/9=1 then tan⁡(α/2)tan⁡(β/2)=

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Ready to Test Your Skills?

free study material

If a, b are the eccentric angles of the
extremities of a focal chord of the ellipse
$x^{2} / 16 + y^{2} / 9 = 1$ then $\tan (α / 2) \tan (β / 2) =$