General point on the ellipse = acost,bsint

 

Eqn of chord of contact to a circle is T=0 which in this case is

 

=> x(acost) + y(asint) - r² = 0

i.e y = (-acost/bsint)x + r²/bsint.........(1)

 

The curve given is

a²x²+b²y²=r⁴

= x² / (r²/a)² + y² / (r²/b)² = 1........(2)

hence it is an ellipse

 

Condition for a line to be a tangent to the ellipse : c² = a²m²+b²

LHS = c² = r⁴/b²sin²t

 

RHS = a²m²+b² = r⁴(a²cos²t)/a²b²cos²t + r⁴/b² = r⁴/b²sin²t = LHS

 

Hence Proved