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 (a) 1 (b)-2 (c) 2 (d) -1

Answer

2

Explanation

2 The equation of to the parabola `y^(2)=4x` having slope m is
`y=mx-2m-m^(3)`
It passes through the point P(h,k). So,
`mh-k-2m-m^(3)=0`
`orm^(3)+(2-h)m+k=0` (1)
which is cubic m has three roots such that the product of roots
`m_(1)m_(2)m_(3)=-k` [From (1)]
But given that `m_(1)m_(2)=alpha.` So,
`m_(3)=-(k)/(alpha)`
But `m_(3)` must satisfy (1). So,
`(-k^(3))/(alpha^(3))+(2-h)(-k)/(alpha)+k=0`
`ork^(2)+2alpha^(2)-halpha^(2)-alpha^(3)=0`
So, the locus of P(h,k) is `y^(2)=alpha^(3)x+(alpha^(3)-2alpha^(2))`.
But given that the locus of is a part of the parabola `y^(2)=4x`.
Therefore, comparing the two, we get
`alpha^(2)=4andalpha^(3)-2alpha^(2)=0`
`:." "alpha=2`