Question

If the area enclosed between the curves \(y^2 = 4kx\) and \(y = kx,k > 0\) is \(\frac{2k}{3}\) sq. units then \(k =\)

• A. \(1\)
• B. \(1.5\)
• C. \(2\)
• D. \(3\)

Hint:

Find the intersection points and integrate the difference of the curves.

Answer 1

The Correct Option is C

Step 1: Integrate with respect to y:
Write $x$ for each curve: parabola $x=\dfrac{y^2}{4k}$, line $x=\dfrac yk$. Intersection at $y=0$ and $y=4$.

Step 2: Area between them:
$A=\int_0^4\left(\dfrac yk-\dfrac{y^2}{4k}\right)dy=\dfrac1k\left[\dfrac{y^2}{2}-\dfrac{y^3}{12}\right]_0^4=\dfrac1k\left(8-\dfrac{16}{3}\right)=\dfrac{8}{3k}$.

Step 3: Solve:
$\dfrac{8}{3k}=\dfrac{2k}{3}$ gives $k=2$. Option C.

Final Answer:
The enclosed area is 8/(3k), and setting it to 2k/3 gives k = 2. \[ \boxed{\text{(C) }2} \]