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)

ankurgupta91

let the equation of one of the normal be
y= mx-2am -am^3
let P(h,k) is the point from which three normals r drawn
nw am^3 +(2a-h)m+k=0
slopes m1,m2,m3 r the roots of this equation
nw m1+m2+m3=0 -----(1)
m1m2+m2m3+m3m1 =(2a-h)/a ------(2)
m1m2m3 = -k/a ---------(3)
and (m1+m2+m3)^2=0
m1^2+m2^2+m3^2 +2( m1m2+m2m3+m3m1) = 0
m1^2 + m2^2 + m3^2 = (2h-4a)/a -----(4)
squaring equation (2)
(m1m2)^2+(m2m3)^2+(m3m1)^2+2[m1m2m3(m1+m2+m3)]
=[2a-h)/a]^2
(m1m2)^2+(m2m3)^2+(m3m1)^2=[2a-h)/a]^2 as (m1+m2+m3=0) (5)

since
tan^-1m1+tan^-1 m2+tan^-1 m3
= tan^-1[(m1+m2+m3 -m1m2m3)/1-(m1m2+m2m3+m3m1)] --(6)
substite the values frm (1)(2)(3)

nd tan^-1 m1^2 +tan-1m2^2 +tan^-1m3^3
= tan^-1[[m1^2+m2^2+m3^3- (m1m2m3)^2]/1-{(m1m2)^2+(m2m3)^2+(m3m1)^2}]
nw substitue value from(3)(4)(5)
nd then equate this value to the equation (6)
u wl get the req locus
hope u gt it
thanks.............

iitkgp_bipin

Equation of normal in slope form :

y = mx - 2am - am³

Let the normals pass through (h,k)

am³ + (2a-h)m + k = 0

m₁ + m₂ + m₃ = 0

m₁m₂ + m₂m₃ + m₃m₁ = (2a-h)/a

m₁m₂m₃ = -k/a

tan^-1m = tan^-1[(m₁+m₂+m_{3 ^-}m₁m₂m₃/(1-m₁m₂ + m₂m₃ + m₃m₁)]

Use above relations to find it.

(m₁ + m₂ + m₃)² = 0

m₁²+m₂²+m₃² +2(m₁m₂ + m₂m₃ + m₃m₁) = 0

m₁²+m₂²+m₃² = -2(2h-a)/a

(m₁m₂ + m₂m₃ + m₃m₁)² = [(2a-h)/a]²

(m₁m₂)² + (m₂m₃)² + (m₃m₁)² + 2m₁m₂m₃(m₁ + m₂ + m₃) = [(2a-h)/a]²

(m₁m₂)² + (m₂m₃)² + (m₃m₁)² = [(2a-h)/a]²

tan^-1m² = tan^-1[(m₁²+m₂²+m₃²-(m₁m₂m₃ )²)/(1-(m₁m₂)²+(m₂m₃)²+(m₃m₁)²)]

Use the above equations to find it.

Equate

tan^-1m and

tan^-1m² which would contain terms in (h,k).

Replace (h,k) by (x,y) to find the required locus.

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