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Question: Tangents PA and PB are drawn to y = x<sup>2</sup>− x + 1 from the point P\(\left( \frac{1}{2},h \rig...

Tangents PA and PB are drawn to y = x2− x + 1 from the point P(12,h)\left( \frac{1}{2},h \right). If the area of triangle PAB is maximum then

A

h = −1/4

B

h = −1/2

C

h = -2

D

None of these

Answer

None of these

Explanation

Solution

Vertex of the given parabola is

Equation of line AB is

12(y+h)=x⋅12−12(x+12)+1\frac { 1 } { 2 } ( y + h ) = x \cdot \frac { 1 } { 2 } - \frac { 1 } { 2 } \left( x + \frac { 1 } { 2 } \right) + 1

⇒ y=32−hy = \frac { 3 } { 2 } - h

Putting this value of y in the equation of parabola we get (x−12)2=(32−h)\left( x - \frac { 1 } { 2 } \right) ^ { 2 } = \left( \frac { 3 } { 2 } - h \right)

⇒ x = 12±34−h\frac { 1 } { 2 } \pm \sqrt { \frac { 3 } { 4 } - h }

⇒

= 2(34−h)3/22 \left( \frac { 3 } { 4 } - h \right) ^ { 3 / 2 }

which is clearly a decreasing function of h.