Sol. We have,

y 2 = a 2 cos 2 x + b 2 sin 2 x

⇒ 2y 2 = a 2 (2 cos 2 x) + b 2 (2 sin 2 x)

⇒ 2y 2 = a 2 (1 + cos 2x) + b 2 (1 – cos 2x)

⇒ 2y 2 = (a 2 + b 2 ) + (a 2 – b 2 ) cos 2x ... (i)

Differentiating with respect to x, we get

4y = –2 (a 2 – b 2 ) + sin 2x

⇒ 2y = – (a 2 – b 2 ) sin 2x ... (ii)

From (i), we have

2y 2 – (a 2 + b 2 ) = (a 2 – b 2 ) cos 2x ... (iii)

squaring (i) and (ii) and adding, we get

4y 2 + {2y 2 –(a 2 + b 2 )} 2 = (a 2 –b 2 ) 2 ⇒ 4y

2 = (a 2 –b 2 ) 2 – (a 2 + b 2 ) 2 ⇒ 4y

2 = –4a 2 b 2 ⇒ + y

2 – (a 2 + b 2 ) = –

Differentiating both sides w.r.t.x, we get

2 + 2y =

⇒ + y =