ax32+by32=1
(32(ax3−1)×a1+32(by3−1)×b1)dxdy=0
(by3−1)×b1)dxdy=−(ax3−1)×a1
dxdy=−ab(byax)−31
= (−ab)32(yx)3−1
let A(x1,y1)⟹m=(−ab)32(yx)3−1
Equation of the tangent is
y−y1=m(x−x1)
y−y1=(−ab)32(yx)3−1(x−x1)
open by cross multiplying
a32x131y−a32x31y1=−b32y131x1+b32y131x1
(b32y131)x+(a32x31)y=x131y131[a32y132+b32x32]
x131y132a32b32[ax132+by132]..................equqation 1
ax32+by32=1
equation 1⟹(b32y131)x+(a32x31)y=x131y132a32b32
x131a32x+y131b32y=1
as per given a=I b=M
I=x131a32
M=y131b32
hence
ax132+by132=1
Comments
Leave a comment