The area in sq. units inside the parabola 5 x2-y=0 but outside the parabola 2 x2-y+9=0 is A. Find (A2/12) .

Question

The area in sq. units inside the parabola 5 x2-y=0 but outside the parabola 2 x2-y+9=0 is A. Find (A2/12) . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/0b5a28e51713001c5c0402aed12a813b-.png" /> Solving y=5 x2 and y=2 x2+9, we get x=± √3 Area, A=∫ limits0√3((2 x2+9)-5 x2) d x =2 ∫ limits0√3(9-3 x2) d x =2[9 x-x3]0√3 =2(9 √3-3 √3)=12 √3 ⇒ (A2/12)=((12 √3)2/12)=36

Q. The area in sq. units inside the parabola $5 x^{2} - y = 0$ but outside the parabola $2 x^{2} - y + 9 = 0$ is A. Find $\frac{A ^{2}}{12} .$

Solving $y = 5 x^{2}$ and
$y = 2 x^{2} + 9,$ we get $x = \pm 3$
Area, $A = 0 \int 3 ((2 x^{2} + 9) - 5 x^{2}) d x$
$= 2 0 \int 3 (9 - 3 x^{2}) d x$
$= 2 [9 x - x^{3}]_{0}^{3}$
$= 2 (93 - 33) = 123$
$\Rightarrow \frac{A ^{2}}{12} = \frac{( 12 3 ) ^{2}}{12} = 36$

Q. The area in sq. units inside the parabola 5x2−y=0 but outside the parabola 2x2−y+9=0 is A. Find 12A2​.

Solving y=5x2 and y=2x2+9, we get x=±3​ Area, A=0∫3​​((2x2+9)−5x2)dx =20∫3​​(9−3x2)dx =2[9x−x3]03​​ =2(93​−33​)=123​ ⇒12A2​=12(123​)2​=36

Q. The area in sq. units inside the parabola $5 x^{2} - y = 0$ but outside the parabola $2 x^{2} - y + 9 = 0$ is A. Find $\frac{A ^{2}}{12} .$

Solving $y = 5 x^{2}$ and
$y = 2 x^{2} + 9,$ we get $x = \pm 3$
Area, $A = 0 \int 3 ((2 x^{2} + 9) - 5 x^{2}) d x$
$= 2 0 \int 3 (9 - 3 x^{2}) d x$
$= 2 [9 x - x^{3}]_{0}^{3}$
$= 2 (93 - 33) = 123$
$\Rightarrow \frac{A ^{2}}{12} = \frac{( 12 3 ) ^{2}}{12} = 36$