Show that the perpendiculars from the centre upon all chords, which join the ends of perpendicular diameters, are of constant length in an ellipse.
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.
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.,.