∵C:(x2+y2−3)+(x2−y2−1)5=0 for point (α,α) α2+α2−3+(α2−α2−1)5=0 ∴α=√2 On differentiating (x2+y2−3)+(x2−y2−1)5=0 we get x+yy′+5(x2−y2−1)4(x−yy′)=0 When x=y=√2 then y′=
3
2
Again on differentiating eq. (i) we get : 1+(y′)2+yy′′+20(x2−y2−1)(2x−2yy′)(x−y′y)+5(x2−y2−1)4(1−y′2−yy′′)=0 For x=y=√2 and y′=