IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

pratik17

find the locus of a point such that slopes m1, m2, m3 of the normals from it to the parabola y^2 = 4ax are related by

tan inverse (m) =

tan inverse (m^2)

pratik17

experts please help......... waitin for an answer.

aneeshrox

note : i [ ]represents inverse
let m1,m2,m3 be the slopes of normals dawn from (x1,y1) on the parabola x^2=4ax
equation of normals : y1=mx1-2am-am^3
from this m1+m2+m3=0
m1m2+m2m3+m3m1= (2a-x1 / a)
m1m2m3= -y1/a
LHS = tan i[ m1]+tan i [m2]+tan i[m3]= pie + tan i[(m1+m2+m3-m1m2m3)/(1-(m1m2+m2m3+m3m1))]
(you can get this by twice applying the formula tan i[] +tan i[]=tan i[(a+b)/1-ab] ...ab>1 in this case)
substituting the values :
=> pie + tan i[(m1+m2+m3-m1m2m3)/(1-(m1m2+m2m3+m3m1))] = -y/(a+x1)
similarly expanding the given RHS
RHS =tan i[ m1^2]+tan i [m2^2]+tan i[m3^2]
=pie +tan i[(m1+m2+m3)^2-(m1m2m3)^2-2(m1m2m3)/1-(m1m2+m2m3+m3m1)^2-2(m1m2+m2m3+m3m1)]
=pie+tan i[(y1^2-2ay1) / (a^2)+(x1^2)+2a.x1]
on equting LHS and RHS you will get the following equation
a(y1-a)=x1(x1-y1)
replacing x1 and y1
ay-a^2=x^2-xy(which is the required locus )
thus this locus represents a parabola

yahiyafirdous

Equation of the normal to the point P(at²,2at) on the parabola, from any point Q(x,y) out side the parabola is given by:

y= -tx +2at+at³

There are three roots of t, hence three normals can be drawn from Q(x,y) whose slopes are the roots of above equation:

Hence:

t₁+t₂+t₃=0....................................(1)

t₁t₂+t₂t₃+t₁t₃=(2a-x)/a....................(2)

t₁t₂t₃=y/a.......................................(3)

_{[ 1]}^{[ 3]} tan^-1(m)=_{[ 1]}^{[ 3]}tan^-1(-t)= -_{[ 1]}^{[ 3]}tan^-1(t) [Since slope= -t]

_{[ 1]}^{[ 3]}tan^-1(t_i) = tan^-1[(t₁+t₂+t₃- t₁t₂t₃)/(1-t₁t₂-t₂t₃-t₁t₃) ]

= -tan^-1[y/(x-a)]................(4)

_{[ 1]}^{[ 3]}tan^-1(t_i²)=tan^-1[{t²₁+t²₂+t²₃- (t₁t₂t₃) ²}/{1-(t₁t₂)²-(t₂t₃)²- (t₁t₃)²} ].....(5)

t²₁+t²₂+t²₃- (t₁t₂t₃) ²= (t₁+t₂+t₃)²- (t₁t₂+t₂t₃+t₁t₃) - (t₁t₂t₃) ²

= -2(2a-x)/a - (y/a)²

1-(t₁t₂)²-(t₂t₃)²- (t₁t₃)²=1 -[(t₁t₂+t₂t₃+t₁t₃)² - 2(t₁t₂²t₃ + t₁t₂t₃² + t₁²t₂t₃)]

=1-[(2a-x)² /a² - 2{t₁t₂t₃(t₂+t₃) +t₁²t₂t₃}]

=1- (2a-x)² /a²=(3a-x)(x-a)/a²

_{[ 1]}^{[ 3]}tan^-1(t_i²)=tan^-1[{-2(2a-x)/a - (y/a)²}/{(3a-x)(x-a)/a²} ]

= - tan^-1[{-2(2a-x)/a - (y/a)²}/{(3a-x)(x-a)/a²} ]..............(6)

Equating equation (4) and (6) and simplify :

y²+3ay-2ax-xy+4a²=0 Ans

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