Slope of line joining the points (0,3) and (5,−2) is ‌
3+2
0−5
=−1 This is equal to the slope of tangent on the curve and that is given by ‌‌
dy
dx
=‌
−c
(x+1)2
⇒‌
dy
dx
=−1 ⇒c=(x+1)2......(i) Equation of line joining the points (0,3) and (5,2) is given by (y−3)=−1(x−0). Solving equation of tangent and the curve for point of intersection ‌
c
x+1
+x=3......(ii) Solving Eqs. (i) and (ii), x=1 On putting this in Eq. (ii), we get c=4