SATHEE: Area Question 2

If the area (in sq units) of the region $\left{(x, y): y^{2} \leq 4 x, x+y \leq 1, \quad x \geq 0, y \geq 0\right}$ is $a \sqrt{2}+b$, then $a-b$ is equal to

(2019 Main, 12 April I) (a) $\frac{10}{3}$ (b) 6 (c) $\frac{8}{3}$ (d) $-\frac{2}{3}$

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Solution:

Given region is $\left{(x, y): y^{2} \leq 4 x, x+y \leq 1, x \geq 0, y \geq 0\right}$

Now, for point $P$, put value of $y=1-x$ to $y^{2}=4 x$, we get $(1-x)^{2}=4 x \Rightarrow x^{2}+1-2 x=4 x$

$$ \begin{aligned} \Rightarrow & x^{2}-6 x+1=0 \ \Rightarrow \quad x & =\frac{6 \pm \sqrt{36-4}}{2} \ & =3 \pm 2 \sqrt{2} . \end{aligned} $$

Since, $x$-coordinate of $P$ less than $x$-coordinate of point $A(1,0)$.

$\therefore \quad x=3-2 \sqrt{2}$

Now, required area

$=\int_{0}^{3-2 \sqrt{2}} 2 \sqrt{x} d x+\int_{3-2 \sqrt{2}}^{1}(1-x) d x$

$=2\left|\frac{x^{3 / 2}}{3 / 2}\right|{0}^{3-2 \sqrt{2}}+x-{\frac{x^{2}}{2}}{3-2 \sqrt{2}}$

$=\frac{4}{3}(3-2 \sqrt{2})^{32}+1-\frac{1}{2}-(3-2 \sqrt{2})+\frac{(3-2 \sqrt{2})^{2}}{2}$

$=\frac{4}{3}\left[(\sqrt{2}-1)^{2}\right]^{3 / 2}+\frac{1}{2}-3+2 \sqrt{2}+\frac{1}{2}(9+8-12 \sqrt{2})$

$=\frac{4}{3}(\sqrt{2}-1)^{3}-\frac{5}{2}+2 \sqrt{2}+\frac{17}{2}-6 \sqrt{2}$

$=\frac{4}{3}(2 \sqrt{2}-3(2)+3(\sqrt{2})-1)-4 \sqrt{2}+6$

$=\frac{4}{3}(5 \sqrt{2}-7)-4 \sqrt{2}+6=\frac{8 \sqrt{2}}{3}-\frac{10}{3}$

$=a \sqrt{2}+b$

So, on comparing $a=\frac{8}{3}$ and $b=-\frac{10}{3}$

(given)

$\therefore \quad a-b=\frac{8}{3}+\frac{10}{3}=6$