Question

If P (t², 2t) t Ī [0, 2] is an arbitrary point on parabola y² = 4x. Q is foot of perpendicular from focus S on the tangent at P, then maximum area of DPQS is-

• A. 1
• B. 2
• C. $\frac { 5 } { 16 }$
• D. 5

Answer 1

Answer: (D)

5

Answer 2

Equation of tangent at P is ty = x + t²

it intersects the line x = 0 at Q

**\**coordinates of Q are (0,t)

\ area of DPQS = $\frac { 1 } { 2 } \left| \begin{array} { c c c } 0 & \mathrm { t } & 1 \\ 1 & 0 & 1 \\ \mathrm { t } ^ { 2 } & 2 \mathrm { t } & 1 \end{array} \right|$

= $\frac { 1 } { 2 }$ [ – t (1 – t²) + 2t] = $\frac { 1 } { 2 }$ (t + t³)

$\frac { 1 } { 2 }$ (3t² + 1) > 0 " t

\ area is maximum for t = 2

Max. area = $\frac { 1 } { 2 }$ [2 + 8] = 5