If the area bounded by the parabola \( x^2+my=0,m\gt 0 \) between its vertex and the latus rectum is \( \displaystyle\frac{3}{2} \) square units, then the value of \( m \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3
\( x^2=-my \) is a parabola with vertex at the origin, opening downwards. Comparing with \( x^2=-4ay \):
\( \displaystyle 4a=m\ \Rightarrow\ a=\frac{m}{4} \)
The latus rectum is the line \( y=-a=-\frac{m}{4} \); it meets the parabola where \( x^2=\frac{m^2}{4} \), i.e. \( x=\pm\frac{m}{2} \).
The region lies between the parabola and this line. By symmetry, take twice the part for \( 0\le x\le\frac{m}{2} \); there the line is below the parabola \( y=-\frac{x^2}{m} \):
\( \displaystyle \text{Area}=2\int_{0}^{m/2}\left(-\frac{x^2}{m}+\frac{m}{4}\right)dx \)
\( \displaystyle =2\left[\frac{mx}{4}-\frac{x^3}{3m}\right]_{0}^{m/2} \)
\( \displaystyle =2\left(\frac{m^2}{8}-\frac{m^2}{24}\right) \)
\( \displaystyle =2\cdot\frac{m^2}{12} \)
\( \displaystyle =\frac{m^2}{6} \)
Since the area is \( \frac{3}{2} \):
\( \displaystyle \frac{m^2}{6}=\frac{3}{2} \)
\( m^2=9 \)
\( m=3 \quad (m\gt 0) \)
Shortcut: The region cut off from a parabola by a chord perpendicular to its axis is \( \frac{2}{3} \) of the enclosing rectangle. Here the chord (latus rectum) has length m and the height is \( \frac{m}{4} \), so the area is
\( \displaystyle \frac{2}{3}\cdot m\cdot\frac{m}{4}=\frac{m^2}{6} \)