1Hence alpha=2
Normals are drawn from the pt P with slopes m_1,m_2,m_3 to the parabola y^2=4x .If Locus of P is a part of the parabola with m_1m_2 = \alpha , find \alpha
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5 Answers
1equation of normal to the parabola y2=4ax with slope m is y=mx-2m-m3
let it pass through P(h,k)
then k=mh-2m-m3
m3+m(2-h)+k=0
hence m1+m2+m3=0
\Sigmam1m2=2-h
m1m2m3=-k
given m1m2=\alpha
m3=-k/\alpha
and m3(m1+m2)+\alpha
=2-h
=-m33 +\alpha
=2-h
-k2/\alpha2 +\alpha=2-h
k2=\alpha2[h-(2-\alpha)]
so locus is y2=\alpha2[x(2-\alpha)]
clearly for \alpha=2
locus of P is a part of parabola whose equation is y2=4ax
1u r rite... [1]
but edit 12th line of ur post... it's not (m33), rather it's (-m32)...... rest is ok... :)