The area in sq- units of the region A - x - y - y 2-2- x - y -4A- 18B- 30- 165- 53-3

A={(x,y):y^22≤ x ≤ y+4} For the point of intersection : y^22=y+4 ⇒ y^2 2y 8=0 ⇒ y=4, 2 A=∫_ 2^4(y+4 y^22)dy ⇒ A=[y^22+4y y^33]^4_ 2 ⇒ A=18 sq. units ...

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Standard XII

Mathematics

Position of a Point W.R.T Parabola

The area in s...

Question

The area (in sq. units) of the region $A = {(x, y) : \frac{y^{2}}{2} \leq x \leq y + 4}$ is :

18

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30

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16

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\frac{53}{3}

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Solution

The correct option is A $18$
$A = {(x, y) : \frac{y^{2}}{2} \leq x \leq y + 4}$
For the point of intersection:
$\frac{y^{2}}{2} = y + 4 \Rightarrow y^{2} - 2 y - 8 = 0$
or, $y = 4, - 2$

$A = 4 \int - 2 (y + 4 - \frac{y^{2}}{2}) d y$
$\Rightarrow A = {[\frac{y^{2}}{2} + 4 y - \frac{y^{3}}{3}]}_{- 2}^{4}$
$\Rightarrow (24 - \frac{32}{3} + 6 - \frac{4}{3})$
or, $A = 18$ sq. units

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The area (in sq. units) of the region A={(x,y):y22≤x≤y+4} is :

The correct option is A 18A={(x,y):y22≤x≤y+4} For the point of intersection: y22=y+4⇒y2−2y−8=0 or, y=4,−2 A=4∫−2(y+4−y22) dy ⇒A=[y22+4y−y33]4−2 ⇒(24−323+6−43) or, A=18 sq. units

The area (in sq. units) of the region $A = {(x, y) : \frac{y^{2}}{2} \leq x \leq y + 4}$ is :