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: Find the intersection:
\(k^2x^2=4kx\) gives \(x=0\) or \(x=\dfrac4k\).

Step 2: Set up the area:
Between these limits the parabola \(y=2\sqrt{kx}\) lies above the line \(y=kx\).
\[ A=\int_0^{4/k}\left(2\sqrt{kx}-kx\right)dx=\left[\dfrac43\sqrt k\,x^{3/2}-\dfrac{kx^2}{2}\right]_0^{4/k} \]

Step 3: Evaluate:
\(\dfrac43\sqrt k\cdot\dfrac{8}{k^{3/2}}-\dfrac k2\cdot\dfrac{16}{k^2}=\dfrac{32}{3k}-\dfrac8k=\dfrac{8}{3k}\).

Step 4: Solve for k:
\(\dfrac{8}{3k}=\dfrac{2k}{3}\), so \(k^2=4\) and \(k=2\) (since \(k>0\)). Option C.

Final Answer:
k = 2. \[ \boxed{\text{(C) }2} \]