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Miscellaneous Exercise 5(II) · Q45

Q.Find the area of the region lying between the parabolas: 4y2=9x4y^2 = 9x and 3x2=16y3x^2 = 16y

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From 4y2=9x4y^2=9x, y=32xy=\frac{3}{2}\sqrt x. From 3x2=16y3x^2=16y, y=316x2y=\frac{3}{16}x^2. Setting them equal via substitution: squaring/substituting leads to x3=64x^3=64, so x=4x=4 and correspondingly y=3y=3. Between x=0x=0 and x=4x=4, the first curve lies above the second (checked at x=1x=1: 1.51.5 vs 0.190.19). So …

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