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Area Under the Curves question

2023 · 25 Jan · Shift 1 · Q45
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  5. /2023 · 25 Jan · Shift 1 · Q45

Area Under the Curves question

2023 · 25 Jan · Shift 1 · Q45

JEE MainMathematicsArea Under the CurvesNumerical+4 / −1
If the area enclosed by the parabolas P1:2y=5x2\mathrm{P_1:2y=5x^2}P1​:2y=5x2 and P2:x2−y+6=0\mathrm{P_2:x^2-y+6=0}P2​:x2−y+6=0 is equal to the area enclosed by P1\mathrm{P_1}P1​ and y=αx,α>0\mathrm{y=\alpha x,\alpha \gt 0}y=αx,α>0, then α3\alpha^3α3 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 600

  1. Write the curves in standard form

Given P1:2y=5x2  ⟹  y=52x2P_1: 2y=5x^2 \implies y=\frac{5}{2}x^2P1​:2y=5x2⟹y=25​x2 and P2:x2−y+6=0  ⟹  y=x2+6.P_2:x^2-y+6=0 \implies y=x^2+6.P2​:x2−y+6=0⟹y=x2+6.

We are told:

  • Area enclosed by P1P_1P1​ and P2P_2P2​
  • equals area enclosed by P1P_1P1​ and the line y=αxy=\alpha xy=αx, where α>0\alpha>0α>0.

We must find α3\alpha^3α3.

  1. Area enclosed by the parabolas P1P_1P1​ and P2P_2P2​

Intersection points are obtained from 52x2=x2+6.\frac{5}{2}x^2=x^2+6.25​x2=x2+6. So, (52−1)x2=6\left(\frac{5}{2}-1\right)x^2=6(25​−1)x2=6 32x2=6\frac{3}{2}x^2=623​x2=6 x2=4  ⟹  x=±2.x^2=4 \implies x=\pm 2.x2=4⟹x=±2.

On [−2,2][-2,2][−2,2], the upper curve is y=x2+6y=x^2+6y=x2+6 and the lower curve is y=52x2.y=\frac{5}{2}x^2.y=25​x2.

Hence area

=\int_{-2}^{2}\left(6-\frac{3}{2}x^2\right)dx.$$ Using symmetry, $$A_1=2\int_{0}^{2}\left(6-\frac{3}{2}x^2\right)dx =2\left[6x-\frac{1}{2}x^3\right]_0^2.$$ Now, $$\left[6x-\frac{1}{2}x^3\right]_0^2=12-4=8.$$ Therefore, $$A_1=2\cdot 8=16.$$ --- 3. **Area enclosed by $P_1$ and the line $y=\alpha x$** Intersection points satisfy $$\frac{5}{2}x^2=\alpha x.$$ So, $$x\left(\frac{5}{2}x-\alpha\right)=0.$$ Thus, $$x=0,\qquad x=\frac{2\alpha}{5}.$$ Since $\alpha>0$, the enclosed region lies between these two intersection points. For $0<x<\frac{2\alpha}{5}$, the line lies above the parabola, so area is $$A_2=\int_0^{2\alpha/5}\left(\alpha x-\frac{5}{2}x^2\right)dx.$$ Compute: $$A_2=\left[\frac{\alpha x^2}{2}-\frac{5}{6}x^3\right]_0^{2\alpha/5}.$$ Substitute $x=\frac{2\alpha}{5}$: First term: $$\frac{\alpha}{2}\left(\frac{2\alpha}{5}\right)^2 =\frac{\alpha}{2}\cdot\frac{4\alpha^2}{25} =\frac{2\alpha^3}{25}.$$ Second term: $$\frac{5}{6}\left(\frac{2\alpha}{5}\right)^3 =\frac{5}{6}\cdot\frac{8\alpha^3}{125} =\frac{4\alpha^3}{75}.$$ Therefore, $$A_2=\frac{2\alpha^3}{25}-\frac{4\alpha^3}{75} =\frac{6\alpha^3-4\alpha^3}{75} =\frac{2\alpha^3}{75}.$$ --- 4. **Equate the two areas** Given $$A_1=A_2,$$ so $$16=\frac{2\alpha^3}{75}.$$ Hence, $$\alpha^3=16\cdot\frac{75}{2}=8\cdot 75=600.$$ --- 5. **Final answer** $$\boxed{\alpha^3=600}$$ The derived answer matches the stored correct answer.
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