Let the equation of parabola having horizontal axis vertex at (h,k) is, (y−k)2=4a(x−h) .....(I) The parabola in equation (I) passes through (−2,1) , (1, 2) and (−1, 3) so, (k−1)2=4a(−2−h) k2−2k+1=−8a−4ah ....(II) Then, (k−2)2=4a(1−h) k2−4k+4=4a−4ah ....(III) And, (3−k)2=4a(−1−h) k2−6k+9=−4a−4ah ....(IV) From equation (II), (III) and (IV), 4a=