The locus of point ‘P’ where the three normals drawn from it on the parabola y 2 = 4ax are such that two of them make complementary angles with x-axis is
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Let parabola y 2 = 4ax
equation of normal y = mx – 2am – am 3 It passes through (h,k)
am
3 + m(2a – h) + k = 0 ….(i)
Let slope of normal are m 1 = tan θ 1
m 2 = tan θ 2
m 3 = tan θ 3
Let m 1 & m 2 make complementary angle it means
θ 2 = 90 0 – θ 1
tan θ 2 = cot θ 1 = 
m 2 = 
from (i)
m 1 × m 2 × m 3 = – k/a
m 3 = – k/a
m 3 is root of equation (i) put value of m 3 in equation (i)
k 2 = a(h – 2a)
the locus of point P is y 2 = a(x – 2a)
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