JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Parabola

The tangent

P T

and the normal

P N

to the parabola

y^{2} = 4 a x

at a point

P

on it meet its axis at points

T

and

N

, respectively. The locus of the centroid of the triangle

P T N

is a parabola whose

Answer: (A), (D)

Solution

Let

P = (a t^{2}, 2 a t)

be a point on the parabola

y^{2} = 4 a x

.

The tangent at

P

t y = x + a t^{2}

, which meets the

x

-axis at

T = (- a t^{2}, 0)

.

The normal at

P

y + t x = 2 a t + a t^{3}

, which meets the

x

-axis at

N = (2 a + a t^{2}, 0)

.

The centroid

G = (h, k)

of triangle

P T N

is:

h = \frac{a t ^{2} - a t ^{2} + 2 a + a t ^{2}}{3} = \frac{2 a + a t ^{2}}{3}, k = \frac{2 a t}{3}

From

k = \frac{2 a t}{3}

, we get

t = \frac{3 k}{2 a}

.

Substituting into

h

h = \frac{2 a + a ( 3 k /2 a ) ^{2}}{3} = \frac{2 a + 9 k ^{2} / ( 4 a )}{3} = \frac{2 a}{3} + \frac{3 k ^{2}}{4 a}

Rearranging:

k^{2} = \frac{4 a}{3} (h - \frac{2 a}{3})

y^{2} = \frac{4 a}{3} (x - \frac{2 a}{3})

.

This is a parabola with vertex

(\frac{2 a}{3}, 0)

and

4 p = \frac{4 a}{3}

, so

p = \frac{a}{3}

.

The focus is at

(\frac{2 a}{3} + \frac{a}{3}, 0) = (a, 0)

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · One or more correct options