If `(a ,b)` is the midpoint of a chord passing through the vertex of the parabola `y^2=4(x+1),` then prove that `2(a+1)=b^2`

Explanation

Since `((2a+1),2b)` lies on `y^(2)=4(x+1)`, we have
`4b^(2)=4(2a+2)`
`or" "b^(2)=2(a+1)`