Curve II y2=3x On differentiating w.r.t. x, we get ‌2yy′‌=3×1 ⇒‌y′‌=‌
3
2y
⇒‌m2‌=‌
3
2y
.....(ii) To get intersection point, x2−4=y2,y2=3x On putting x2−4=3x, ‌x2−3x−4=0 ‌⇒‌‌x2−4x+x−4=0 ‌⇒‌‌x(x−4)+1(x−4)=0‌. ‌ ‌⇒‌‌(x−4)(x+1)=0 ‌x=4,−1 From the graph it is clear that x=4
∴‌‌y2=3×4 ⇒‌‌y=±2√3 At point (4,2√3) from Eqs. (i) and (ii), we get m1‌=‌