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Answer

`((3)/(4))`

Explanation

We know that, normal for `y^(2) = 4ax` is given by, `y = mx - 2am - am^(3)`
`:.` Equation of normal for `y^(2) = x` is
`y = mx - (m)/(2) - (m^(3))/(4) = 0`
`[:' a = (1)/(4)]`
since normal passes through (c,0)
`:.mc - (m)/(2) - (m^(3))/(4) = 0`
`implies m (c - (1)/(2) - (m^(2))/(4)) = 0 implies m = 0`
or `m^(2) = 4 (c - (1)/(2))`
`implies m = 0`, the equation of normal is y = 0
Also, `m^(2) ge 0`
`implies c - 1//2 ge 0 implies c ge 1//3`
At `c = 1//2 implies m = 0`
Now, for other normals tobe perpendicular to each other, we must have `m_(1).m_(2) = -`
or `(m^(2))/(4) + ((1)/(2) - c) = `, has `m_(1) m_(2) = -1`
`implies (((1)/(2) - C))/(1//4) = - 1 implies (1)/(2) - c = - (1)/(4)`
`implies c = (3)/(4)`