The area of the region bounded by the parabola y² = 4kx, where k > 0 and its latus rectum is 24 square units. What is the value of k ?

Calculation:

The equation of parabola is y² = 4kx      ------(1)

Let O be the vertex, S be the focus, and LL' be the latus rectum of the parabola.

The equation of latus rectum is x = k.

Also, here the parabola is symmetric about x−axis.

Required area, A = 2(area of OSL)

⇒ Required area, A = 2 \(\displaystyle \int_0^ky\ dx\)

⇒ A = 2 \(\displaystyle \int_0^k2√ k\ √ x\ dx\)

⇒ A = 2.2√k \(\displaystyle \int_0^k \ x^{\frac{1}{2}} dx\)

⇒ A = 4√k \(\displaystyle \ \left [ \frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^k \)

⇒ A = \(\displaystyle \frac{8}{3}\ \sqrt k\ [k^\frac{3}{2}-0^\frac{3}{2}]\)

⇒ A = \(\displaystyle \frac{8}{3}\ k^2\)

Since the area of the given parabola is 24.

⇒ 24 = \(\displaystyle \frac{8}{3}\ k^2\)

⇒ k = ± 3

As the value of  k > 0, k = 3.

∴ The value of k = 3.