If the area bounded by the parabola x^2+my=0, m>0 between its vertex and the latus rectum is 3/2 square units, then the value of m is Option 1: 1 Option 2: 2 Option 3: 3 Option 4: 4

Answer: 3. Solution: The parabola is x^2 = -my with 4a = m, so the latus rectum is y = -m/4, meeting the curve at x = ±m/2. Area = 2_0^(m/2)(m/4-x^2/m)dx = m^2/6. Setting m^2/6 = 3/2 gives m^2 = 9, so m = 3.

Calculus

Application of Integrals

Hard

core

If the area bounded by the parabola $x^2+my=0, m>0$ between its vertex and the latus rectum is $\dfrac{3}{2}$ square units, then the value of $m$ is

✅ Correct Option: 3

The parabola is $x^2 = -my$ with $4a = m$ , so the latus rectum is $y = -\dfrac{m}{4}$ , meeting the curve at $x = \pm\dfrac{m}{2}$ .

Area $= 2\int_0^{m/2}\left(\dfrac{m}{4}-\dfrac{x^2}{m}\right)dx = \dfrac{m^2}{6}$ .

Setting $\dfrac{m^2}{6} = \dfrac{3}{2}$ gives $m^2 = 9$ , so $m = 3$ .

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If the area bounded by the parabola $x^2+my=0, m>0$ between its vertex and the latus rectum is $\dfrac{3}{2}$ square units, then the value of $m$ is