Question

Area bounded by the parabola y = x²− 2x + 3 and tangents drawn to it from the point P(1, 0) is equal to

• A. $4 \sqrt { 2 }$sq. units
• B. $\frac { 4 \sqrt { 2 } } { 3 }$sq. units
• C. $\frac { 8 \sqrt { 2 } } { 3 }$sq. units
• D. $\frac { 16 } { 3 } \sqrt { 2 }$sq. units

Answer 1

Answer: (C)

$\frac { 8 \sqrt { 2 } } { 3 }$sq. units

Answer 2

Let the drawn tangents be PA and PB. AB is clearly the chord of contact of point 'P', thus equation of AB is $\frac { 1 } { 2 }$.(y + 0) = x.1 − (x+ 1) + 3 i.e. y = 4

x coordinates of points A and B will be given by, x² − 2x + 3 = 4 i.e. x² − 2x − 1 = 0

⇒ x = 1 ± $\sqrt { 2 }$

Thus AB = $2 \sqrt { 2 }$ units.

Hence ∆_PAB = $\frac { 1 } { 2 } ( 2 \sqrt { 2 } ) \cdot 4 = 4 \sqrt { 2 }$

Now area bounded by line AB and parabola is equal to $\int _ { 1 - \sqrt { 2 } } ^ { 1 + \sqrt { 2 } } \left( 4 - \left( x ^ { 2 } - 2 x + 3 \right) \right) d x$ i.e equal to $\frac { 4 \sqrt { 2 } } { 3 }$ sq. units.

Thus required area = $4 \sqrt { 2 } - \frac { 4 \sqrt { 2 } } { 3 } = \frac { 8 \sqrt { 2 } } { 3 }$ sq. units.