Normals are drawn from a point P with slopes `m_1,m_2 and m_3` are drawn from the point p not from the parabola `y^2=4x`. For `m_1m_2=alpha`, if the locus of the point P is a part of the parabola itself, then the value of `alpha` is

Options

A. 1
B. -2
C. 2
D. -1

Answer

2

Explanation

We know equation of normal to `y^(2) = 4ax` is
`y = mx - 2am - am^(3)`
Thus, equation of normal to `y^(2) = 4x` is
`y = mx - 2m - m^(3)` let it passes through (h,k)
`implies k = mh - 2m - m^(3)`
or `m^(3) + m(2 - h) + k = 0`
Here, `m_(1) + m_(2) + m_(3) = 0`
`m_(1)m_(2) + m_(2)m_(3) + m_(3)m_(1) = 2 - h`
`m_(1)m_(2)m_(3) =- k` where `m_(1)m_(2) = alha`
`implies m_(3) = - (k)/(alpha)` it must satisfy Eq. (i)
`implies - (k^(3))/(alpha^(3)) - (k)/(alpha) (2 - k) + k = 0`
`implies k^(2) = alpha h 2alpha^(2) + alpha^(3)`
`implies y^(2) = alpha^(2) x - 2alpha^(2 + alpha^(3)`
On comparing with `y^(2 ) = 4x`
`implies alpha^(2) = 4`
and `-2alpha^(2) + alpha^(3) = 0`
`implies alpha = 2`