Let a curve y=y(x) pass through the point (3,3) and the area of the region under this curve, above the x-axis and between the abscissae 3 and x(>3) be (y/x)³. If this curve also passes through the point (α, 6 √10) in the first quadrant, then α is equal to ___________.

Solution

Maths sir · 4 steps

Area given as a function of x? Differentiate both sides, and the integral turns into a differential equation.

1. Turn area into ODE. The condition is ∫₃ˣ y dt = y³/x³. Differentiate both sides with respect to x, using the quotient rule on the right: y = (3y² y')/x³ - 3y³/x⁴. Multiply by x⁴ and divide by y to get 3xyy' = x⁴ + 3y².
2. Substitute v equals y². Put v = y², so v' = 2yy' and the equation becomes 3x/2v' = x⁴ + 3v. Rearranged, this is the linear equation v' - 2/xv = 2/3x³. Its integrating factor is 1/x².

   d/dx(y²/x²) = 2/3x ⇒ y²/x² = x²/3 + C

3. Use the point. At (3,3) the equation gives 1 = 3 + C, so C = -2. The curve is y² = x⁴/3 - 2x².
4. So. Put y = 6√10, so y² = 360: α⁴ - 6α² - 1080 = 0. This gives α² = (6 + 66)/2 = 36, rejecting the negative root. In the first quadrant α > 0, so α = 6.

The idea to keep. When an area or integral up to x is given in terms of y, differentiate first. If yy' appears next to y², put v = y² to make the equation linear.

Answer from the NTA final answer key. Solution by PrepWiser.