Question

The locus of point ‘P’ where the three normals drawn from it on the parabola y² = 4ax are such that two of them make complementary angles with x-axis is

• A. y² = a(x – 3a)
• B. y² = a (x – 2a)
• C. y² = a (x – a)
• D. y*²* = a (x – 4a)

Answer 1

Answer: (B)

y² = a (x – 2a)

Answer 2

Let parabola y² = 4ax

equation of normal y = mx – 2am – am³

It passes through (h,k)

am³ + m(2a – h) + k = 0 ….(i)

Let slope of normal are m₁ = tanq₁

m₂ = tanq₂

m₃ = tanq₃

Let m₁ & m₂ make complementary angle it means

q₂ = 90⁰ – q₁

tanq₂ = cot q₁ = $\frac{1}{\tan\theta_{1}}$

m₂ = $\frac{1}{m_{1}}$

from (i)

m₁ × m₂ × m₃ = – k/a

m₃ = – k/a

m₃ is root of equation (i) put value of m₃ in equation (i)

k² = a(h – 2a)

the locus of point P is y² = a(x – 2a)