JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Parabola

Let

P

and

Q

be distinct points on the parabola

y^{2} = 2 x

such that a circle with

P Q

as diameter passes through the vertex

O

of the parabola. If

P

lies in the first quadrant and the area of the triangle

△ O P Q

32

, then which of the following is (are) the coordinates of

P

Answer: (A), (D)

Solution

Let

P = (2 t_{1}^{2}, 2 t_{1})

and

Q = (2 t_{2}^{2}, 2 t_{2})

be points on

y^{2} = 2 x

.

Since the circle with diameter

P Q

passes through the origin

O

, we have

O P ⊥ O Q

O P \cdot O Q = 0

2 t_{1}^{2} \cdot 2 t_{2}^{2} + 2 t_{1} \cdot 2 t_{2} = 0

4 t_{1}^{2} t_{2}^{2} + 4 t_{1} t_{2} = 0

t_{1} t_{2} (t_{1} t_{2} + 1) = 0

Since

P

and

Q

are distinct and not at the origin,

t_{1} t_{2} = - 1

.

Area of

△ O P Q

\frac{1}{2} ∣ x_{P} y_{Q} - x_{Q} y_{P} ∣ = \frac{1}{2} ∣2 t_{1}^{2} \cdot 2 t_{2} - 2 t_{2}^{2} \cdot 2 t_{1} ∣ = 2∣ t_{1} t_{2} ∣∣ t_{1} - t_{2} ∣ = 2∣ t_{1} - t_{2} ∣

Given area is

32

2∣ t_{1} - t_{2} ∣ = 32

∣ t_{1} - t_{2} ∣ = \frac{3 2}{2}

With

t_{1} t_{2} = - 1

and

t_{1} > 0

(first quadrant), let

t_{2} = - \frac{1}{t _{1}}

t_{1} + \frac{1}{t _{1}} = \frac{3 2}{2}

t_{1} + \frac{1}{t _{1}} = \frac{3 2}{2}

2 t_{1}^{2} - 32 t_{1} + 2 = 0

t_{1} = \frac{3 2 \pm 18 - 16}{4} = \frac{3 2 \pm 2}{4}

t_{1} = 2 or \frac{2}{2}

For

t_{1} = 2

P = (2 \cdot 2, 22) = (4, 22)

✓ (option A)

For

t_{1} = \frac{2}{2}

P = (2 \cdot \frac{1}{2}, 2 \cdot \frac{2}{2}) = (1, 2)

✓ (option D)

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · One or more correct options