Question

84) The locus of poles of normal chords of the parabola y2 = 4ax is
1) (x + 2a) y2 = 4a3 2) (x + 2a) y2 + 4a3 = 0 3) (x – 2a) y2 = 4a3 4) (x – 2a) y2 + 4a3 = 0

Answer 1

Given:

Parabola y^2 = 4 a x

To find:

The locus of poles of normal chords

Solution:

By formula,

Normal is given by,

y = m x - 2 a am - a m^3

Let the pole be ( x1, y1 ),

y y1 = 2a ( x + x1 )

Comparing the coefficients,

We get,

y1 = 2 a / m = 2 a x1 / - 2 a m - a am^3

Hence,

m = 2 a / y1

I = a m^2 + 2 a + x1 = 0

Substituting,

We get,

( x + 2 a ) y^2 + 4 a^3 = 0

Hence, ( x + 2 a ) y^2 + 4 a^3 = 0 is the  locus of poles of normal chords of the parabola y^2 = 4 a x