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

A line

L : y = m x + 3

meets

y

-axis at

E (0, 3)

and the arc of the parabola

y^{2} = 16 x

0 \leq y \leq 6

at the point

F (x_{0}, y_{0})

. The tangent to the parabola at

F (x_{0}, y_{0})

intersects the

y

-axis at

G (0, y_{1})

. The slope

m

of the line

L

is chosen such that the area of the triangle

E F G

has a local maximum.
Match List I with List II and select the correct answer using the code given below the lists :

Answer: (A)

Solution

Let

F = (4 t^{2}, 8 t)

be a point on the parabola

y^{2} = 16 x

(parametric form).

The tangent at

F

t y = x + 4 t^{2}

, which meets the

y

-axis at

G (0, 4 t)

.

So

E = (0, 3)

F = (4 t^{2}, 8 t)

G = (0, 4 t)

.

Area of

△ E F G = \frac{1}{2} ∣ x_{F} (y_{E} - y_{G}) ∣ = \frac{1}{2} ∣4 t^{2} (3 - 4 t) ∣ = 2 t^{2} ∣3 - 4 t ∣

For

0 \leq y \leq 6

, we have

0 \leq t \leq 3/4

.

Area

= 2 t^{2} (3 - 4 t) = 6 t^{2} - 8 t^{3}

for

t \in [0, 3/4]

\frac{d A}{d t} = 12 t - 24 t^{2} = 12 t (1 - 2 t) = 0

This gives

t = 0

t = 1/2

.

At

t = 1/2

F = (1, 4)

G = (0, 2)

, Area

= 6 (1/4) - 8 (1/8) = 3/2 - 1 = 1/2

Slope

m = \frac{4 - 3}{1 - 0} = 1

So:

m = 1

(matches 4), Maximum area

= 1/2

(matches 1),

y_{0} = 4

(matches 2),

y_{1} = 2

(matches 3)

Answer: P-4, Q-1, R-2, S-3

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