l, m be the portions of the axes of x and y intercepted by the tangent at any point ( x,y) on the curve
(ax)32+(by)32=1
b32x32+a32y32=a32b32 ••••equation(1)
The parametric coordinate of the point (x,y) is (acos3θ,bsin3θ) , θ is a parameter.
By the problem, equation of tangent is
lx+my=1 ••••equation(2) to the curve at ( acos³θ , bsin³θ)
Now we find equation of tangent at (acos³θ , bsin³θ)
Differentiating equation (1) with respect to x we get
32b32x−31+32a32y−31dxdy=0
=> dxdy=−a32x31b32y31
Now [dxdy](acos3θ,bsin3θ)=−a32(acos3θ)31b32(bsin3θ)31=−abtanθ
So equation of tangent at (acos3θ,bsin3θ) will be
y−bsin3θ =−abtanθ(x−acos3θ)
=> (acosθ)y+(bsinθ)x=absinθcos3θ+abcosθsin3θ
=> (bsinθ)x+(acosθ)y=absinθcosθ(cos2θ+sin2θ)
=> (bsinθ)x+(acosθ)y=absinθcosθ
=> absinθcosθ(bsinθ)x+(acosθ)y=1
=> acosθx+bsinθy=1 ••••equation(3)
Obviously equation (2) and equation (3) are identical.
So equating coefficients we get
acosθl=bsinθm=11
=> l = acosθ and m = bsinθ
=> al=cosθ and bm=sinθ
So (al)2+(bm)2=cos2θ+sin2θ
=> (al)2+(bm)2=1
=> (a2l2)+(b2m2)=1
Hence proved.
Comments