The correct options are
A 2 if a=5, b=3
C 2 if a=6, b=4
Let P(acosθ,bsinθ) be a variable point on x2a2+y2b2=1
Therefore, 2xa2+2yb2dydx=0
dydx∣∣p=−baxy=−bacotθ
∴ Equation of normal at P
⇒y−bsinθ=−1bacotθ(x−acosθ)
⇒y−bsinθ=abtanθ⋅(x−acosθ)
⇒axsinθ−a2sinθcosθ=b ycosθ−b2sinθcosθ
⇒axsinθ−b ycosθ=(a2−b2)sinθcosθ
⇒axsecθ−b y cosecθ=(a2−b2)
Centre of standard ellipse is origin.
Length of perpendicular from ′O′ on the normal at ′P′ is
l=a2−b2√a2sec2θ+b2 cosec2θ.....(i)
For l to be maximum then denominator should be minimum.
Let Z=a2sec2θ+b2 cosec2θ...(ii)
So, dZdθ=0⇒2a2sec2θtanθ+2b2 cosec2θcotθ=0
∴tan4θ=b2a2⇒tan2θ=ba⇒tanθ=±√ba
For tan2θ=ba, Put in equation (ii)
Z=a2(1+ba)+b2(1+ab)
⇒Z=a2+2ab+b2=(a+b)2
Put minimum value of Z in equation (i)
∴lmax=a2−b2a+b=a−b