Let P≡(x1,y1), Q≡(x2,y2)
y=x3dydx=3x2⇒m1=3x21 ⋯(1)
Equation of tangent at P(x1,y1) is
y−y1=3x21(x−x1)⇒y−x31=3x21(x−x1)⇒y−3x21x+2x31=0 ⋯(2)
Tangent at P intersects the curve again at Q(x2,y2)
So, substituting (x2,x32) in eqn(2), we get
x32−3x21x2+2x31=0⇒(x2−x1)(x22+x1x2+x21−3x21)=0
⇒x22+x1x2−2x21=0 (∵x1≠x2)
⇒(x2−x1)(x2+2x1)=0
⇒x2=−2x1 (∵x1≠x2)
Now, slope of tangent at Q(x2,y2) is
m2=3x22=3(−2x1)2=4(3x21)⇒k=4