Question

If the area of the region bounded by the parabola \(y^2 = 4kx\) and the line \(x = k\), (where \(k > 0\)) is \(\frac{128}{3}\) sq. units, then the value of \(sin^{-1}(\frac{2}{k})\) is equal to...

• A. \(\frac{π}{4}\)
• B. \(\frac{π}{6}\)
• C. \(\frac{π}{3}\)
• D. \(\frac{π}{2}\)

Hint:

Compute the area symmetric about the x-axis and solve for k.

Answer 1

The Correct Option is B

Step 1: Integrate Along y:
Use horizontal strips. At height $y$ the strip extends from $x=\dfrac{y^2}{4k}$ to $x=k$. The line meets the parabola at $y=\pm2k$.

Step 2: Area:
\[ A=\int_{-2k}^{2k}\left(k-\frac{y^2}{4k}\right)dy=2\left[2k^2-\frac{8k^3}{12k}\right]=2\left[2k^2-\frac{2k^2}3\right]=\frac{8k^2}3 \]

Step 3: Result:
Setting $\dfrac{8k^2}{3}=\dfrac{128}{3}$ gives $k=4$, so $\sin^{-1}(2/4)=\dfrac\pi6$. Option (B).

Final Answer:
Option (B). \[ \boxed{\text{(B) } \frac{\pi}{6}} \]