If the area (in sq.

Question

If the area (in sq. units) of the region $\{(x,y) : y^2\le4x,\ x+y\le1,\ x\ge0,\ y\ge0\}$ is $a\sqrt2+b$, then $a-b$ is equal to:

Accepted Answer

Let P be the point common to x + y = 1 and y2 = 4x

Then y2 = 4(1-y) $$ \Rightarrow $$ y2 - 4y - 4 = 0

$$ \Rightarrow y = {{ - 4 \pm \sqrt {16 + 16} } \over 2}$$

$$ \Rightarrow y =  - 2 + 2\sqrt 2 $$

Then P is (3 - $$ 2\sqrt 2$$, -2 + $$ 2\sqrt 2$$).

Hence shaded area = Area of region (OPN) + area of ($$\Delta OPQ$$)

$$ \Rightarrow {\int\limits_0^{3 - 2\sqrt 2 } {2\sqrt {xdx}  + {1 \over 2}\left[ {1 - (3 - 2\sqrt 2 )} \right]} ^2}$$

$$ \Rightarrow {2 \over 3}2\left( {\sqrt 2  - 1} \right)\left( {3 - 2\sqrt 2 } \right) + {1 \over 2}{\left[ {2(\sqrt 2  - 1)} \right]^2}$$

$$ \Rightarrow {4 \over 3}\left\{ { - 7 + 5\sqrt 2 } \right\} + 2\left( {3 - 2\sqrt 2 } \right)$$

$$ \Rightarrow \left( {{{20} \over 3} - 4} \right)\sqrt 2  + 6 - {{28} \over 3}$$

$$ \Rightarrow {8 \over 3}\sqrt 2  - {{10} \over 3}$$

Then a = $${8 \over 3}$$ and b = $${-10 \over 3}$$, so a - b = 6

Solutions · 1