The area (in sq. units) of the region bounded by the curve \( x=at^2,y=2at \) between the ordinates corresponding to t = 1 and t = 2 is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{56}{3}a^2 \)
Eliminating t, \( t=\frac{y}{2a} \) gives \( x=a\cdot\frac{y^2}{4a^2} \), that is, the parabola \( y^2=4ax \), which is symmetric about the x-axis.
The ordinates are the vertical lines at t = 1 and t = 2: \( x=a \) and \( x=4a \). The region bounded by the parabola and these two lines lies on both sides of the x-axis, so its area is twice the area of the part above the axis.
Above the axis \( y=2at \) and \( dx=2at\,dt \); as x goes from a to 4a, t goes from 1 to 2:
\( \displaystyle \int_{a}^{4a}y\,dx=\int_{1}^{2}2at\cdot 2at\,dt \)
\( \displaystyle =4a^2\int_{1}^{2}t^2\,dt \)
\( \displaystyle =4a^2\left[\frac{t^3}{3}\right]_{1}^{2} \)
\( \displaystyle =4a^2\cdot\frac{7}{3} \)
\( \displaystyle =\frac{28}{3}a^2 \)
\( \displaystyle \text{Area}=2\times\frac{28}{3}a^2=\frac{56}{3}a^2 \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{14}{3}a^2 \) — Incorrect. The area works out to \( \frac{56}{3}a^2 \), not \( \frac{14}{3}a^2 \).
Option (2) \( \displaystyle\frac{28}{3}a^2 \) — Incorrect. \( \frac{28}{3}a^2 \) is only the part above the x-axis; the parabola also bounds the region below the axis.
Option (4) \( \displaystyle\frac{64}{3}a^2 \) — Incorrect. \( \frac{64}{3}a^2 \) is the area from x = 0 to x = 4a on both sides; the region starts at x = a, not at the vertex.
NCERT Exemplar:
Application of Integrals
