KCET 2016 Math Solved Paper

Given curves,→
x3−3xy2+2=0→ (1)
3x2y−y3=2→ (2)
Differentiating Eqs. (1) and (2) both the sides with respect to x, we get
3x2−3(y2+x(2yy′))=0
⇒x2=y2+2xyy′
⇒y′=‌

x2−y2

2xy

=m1 (Let) → (3)
3(x2y′+2xy)3y2y′=0
⇒x2y′+2xy−y2y′=0
⇒y′=‌

−2xy

x2−y2

=m2 (Let) → (4)
From Eqs. (3) and (4), we get
m1m2=−1
Hence, the two curves cut at an angle

π

2

i.e. right angle