71Use the formula for chord of ellipse joining two points with eccentric angles α , α + Π/2
You'll get the equation of chord as
x/a cos(α+Π/4) + y/b sin(α+Π/4) = cos (Π/4)
Obviously, perpendicular on it from (0,0) are of constant length.,.
Show that the perpendiculars from the centre upon all chords, which join the ends of perpendicular diameters, are of constant length in an ellipse.
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3 Answers
262using the standard eqn of ellipse -> x2/a2 +y2/b2 =1
let one the diameters be y=mx
therefore the other diameter is y =-b2a2mx
( since m1*m2 = -b2/a2)
find out their points of intersection with the ellipse.
then form the eqn of a chord and find the perp. distance.
OR you can also use parametric coordinates (acost,bsint) and (acos(90+t),bsin(90+t) )as the ends of diameters and then proceed.
1I was looking for a less tedious method to solve it. And is m1*m2 not equal to -1 as they are pendiculars.
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