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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 |
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Answer» Correct Answer - 2 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` |
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