JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Parabola

Let

P Q

be a focal chord of the parabola

y^{2} = 4 a x

. The tangents to the parabola at

P

and

Q

meet at a point lying on the line

y = 2 x + a, a > 0

.

If chord

P Q

subtends an angle

θ

at the vertex of

y^{2} = 4 a x

, then

tan θ =

Answer: (D)

Solution

From the previous question, the chord

P Q

has equation

2 x + y - 2 a = 0

, or

y = - 2 x + 2 a

.

This chord passes through the focus

(a, 0)

and has slope

- 2

.

To find the angle subtended at the vertex (origin), we need the slopes of

O P

and

O Q

.

Substituting

y^{2} = 4 a x

into

y = - 2 x + 2 a

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

y^{2} + 2 a y - 4 a^{2} = 0

Let the two

y

-values be

y_{1}

and

y_{2}

. Then:

The slopes from origin are

m_{1} = y_{1} / x_{1}

and

m_{2} = y_{2} / x_{2}

.

Since

y^{2} = 4 a x

, we have

x = y^{2} / (4 a)

, so

m = y / x = 4 a / y

.

Thus

m_{1} m_{2} = \frac{16 a ^{2}}{y _{1} y _{2}} = \frac{16 a ^{2}}{- 4 a ^{2}} = - 4

And

m_{1} + m_{2} = \frac{4 a}{y _{1}} + \frac{4 a}{y _{2}} = 4 a \cdot \frac{y _{1} + y _{2}}{y _{1} y _{2}} = 4 a \cdot \frac{- 2 a}{- 4 a ^{2}} = 2

tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}} = \frac{\pm ( m _{1} + m _{2} ) ^{2} - 4 m _{1} m _{2}}{1 + m _{1} m _{2}} = \frac{\pm 4 + 16}{1 - 4} = \frac{\pm 2 5}{- 3} = \mp \frac{2 5}{3}

Taking the negative value:

tan θ = - \frac{2 5}{3}

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Single correct option