A man starts walking from the point $\mathrm{P}(-3,4)$, touches the x -axis at R , and then turns to reach at the point $\mathrm{Q}(0,2)$. The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50(P R)^{2}+(R Q)^{2}$ is equal to

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Straight Lines

A man starts walking from the point

P (- 3, 4)

, touches the x -axis at R , and then turns to reach at the point

Q (0, 2)

. The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then

50 (P R)^{2} + (R Q)^{2}

is equal to

Answer: 1250

For minimum time at constant speed, the path length

P R + R Q

must be minimized.

By the reflection principle, reflect

P (- 3, 4)

across the x-axis to get

P^{'} (- 3, - 4)

. The shortest path from

P

R

Q

equals the straight-line distance from

P^{'}

Q

.

Let

R = (r, 0)

be the point on the x-axis. The line

P^{'} Q

intersects the x-axis at the optimal

R

.

Line through

P^{'} (- 3, - 4)

and

Q (0, 2)

: slope

= \frac{2 - ( - 4 )}{0 - ( - 3 )} = \frac{6}{3} = 2

Equation:

y - 2 = 2 (x - 0) \Rightarrow y = 2 x + 2

At x-axis (

y = 0

0 = 2 x + 2 \Rightarrow x = - 1

, so

R = (- 1, 0)

P R^{2} = (- 1 - (- 3))^{2} + (0 - 4)^{2} = 4 + 16 = 20

R Q^{2} = (0 - (- 1))^{2} + (2 - 0)^{2} = 1 + 4 = 5

50 (P R)^{2} + (R Q)^{2} = 50 (20) + 5 = 1000 + 250 = 1250

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Integer answer