EQUATION OF A CHORD JOINING θθ and ϕϕ:Consider two points θandϕθandϕ lying on the ellipse x2a2+y2b2=1.x2a2+y2b2=1.
We wish to determine the equation of the chord joining these two points.

Using the two point form, we have
y−bsinθx−acosθ=bsinθ−bsinϕacosθ−acosϕ⇒y−bsinθx−acosθ=−bacos(θ+ϕ2)sin(θ+ϕ2)⇒xacos(θ+ϕ2)+ybsin(θ+ϕ2)=cosθcos(θ+ϕ2)+sinθsin(θ+ϕ2)⇒xacos(θ+ϕ2)+ybsin(θ+ϕ2)=cos(θ−ϕ2)y−bsinθx−acosθ=bsinθ−bsinϕacosθ−acosϕ⇒y−bsinθx−acosθ=−bacos(θ+ϕ2)sin(θ+ϕ2)⇒xacos(θ+ϕ2)+ybsin(θ+ϕ2)=cosθcos(θ+ϕ2)+sinθsin(θ+ϕ2)⇒xacos(θ+ϕ2)+ybsin(θ+ϕ2)=cos(θ−ϕ2)
This is the most general equation of a chord joining any two arbitrary points θandϕθandϕ on the ellipse. As an exercise using this form try writing
(a) the equation of any chord passing through the origin
and (b) the equation of the latus-recta by using the eccentric angles of its extremities which we derived earlier.
Example – 9
Suppose that the chord joining the points θ1andθ2θ1andθ2 on the ellipse x2a2+y2b2=1x2a2+y2b2=1 intersects the major-axis in (h, 0). Show that
tan(θ12)tan(θ22)=h−ah+a.tan(θ12)tan(θ22)=h−ah+a.
Solution: By the result we just derived, the equation of the chord joining θ1andθ2θ1andθ2 is
xacos(θ1+θ22)+ybsin(θ1+θ22)=cos(θ1−θ22)xacos(θ1+θ22)+ybsin(θ1+θ22)=cos(θ1−θ22)
If this passes through (h, 0), we have
−hacos(θ1+θ22)=cos(θ1−θ22)−hacos(θ1+θ22)=cos(θ1−θ22)
⇒ha=cos(θ1−θ22)cos(θ1+θ22)⇒h−ah+a=cos(θ1−θ22)−cos(θ1+θ22)cos(θ1−θ22)+cos(θ1+θ22)=tan(θ12)tan(θ22)⇒ha=cos(θ1−θ22)cos(θ1+θ22)⇒h−ah+a=cos(θ1−θ22)−cos(θ1+θ22)cos(θ1−θ22)+cos(θ1+θ22)=tan(θ12)tan(θ22)