Understanding the Parabola and its Latus Rectum

The given equation of the parabola is \(y^2 = 4kx\), where \(k > 0\). This is a standard form of a parabola that opens to the right, with its vertex at the origin \((0,0)\) and its axis along the positive x-axis.

For a standard parabola \(y^2 = 4ax\), the focus is at \((a, 0)\) and the equation of the latus rectum is \(x = a\). Comparing \(y^2 = 4kx\) with \(y^2 = 4ax\), we see that \(a=k\). Therefore, for the given parabola \(y^2 = 4kx\), the focus is at \((k, 0)\) and the equation of the latus rectum is \(x = k\).

The latus rectum is a line segment perpendicular to the axis of the parabola, passing through the focus and having its endpoints on the parabola. The region bounded by the parabola \(y^2 = 4kx\) and its latus rectum \(x=k\) is the area enclosed by the curve \(y^2 = 4kx\) and the vertical line \(x=k\) from \(x=0\) to \(x=k\).

Calculating the Area Bounded by the Parabola and Latus Rectum

The region is symmetric about the x-axis. We can calculate the area of the upper half (where \(y \ge 0\)) and double it to get the total area. From the equation \(y^2 = 4kx\), for \(y \ge 0\), we have \(y = \sqrt{4kx} = 2\sqrt{k}\sqrt{x}\).

The area \(A\) is given by the integral of \(y\) with respect to \(x\) from \(x=0\) to \(x=k\), multiplied by 2 due to symmetry:

$$A = 2 \int_{0}^{k} y \, dx = 2 \int_{0}^{k} 2\sqrt{k}\sqrt{x} \, dx$$

$$A = 4\sqrt{k} \int_{0}^{k} x^{1/2} \, dx$$

Now, we integrate \(x^{1/2}\) with respect to \(x\):

$$\int x^{1/2} \, dx = \frac{x^{1/2 + 1}}{1/2 + 1} + C = \frac{x^{3/2}}{3/2} + C = \frac{2}{3}x^{3/2} + C$$

Applying the definite integral limits:

$$A = 4\sqrt{k} \left[ \frac{2}{3} x^{3/2} \right]_{0}^{k}$$

$$A = 4\sqrt{k} \left( \frac{2}{3} k^{3/2} - \frac{2}{3} 0^{3/2} \right)$$

$$A = 4\sqrt{k} \left( \frac{2}{3} k^{3/2} \right)$$

$$A = \frac{8}{3} k^{1/2} k^{3/2}$$

Using the property \(a^m \cdot a^n = a^{m+n}\):

$$A = \frac{8}{3} k^{1/2 + 3/2} = \frac{8}{3} k^{4/2} = \frac{8}{3} k^2$$

The area bounded by the parabola \(y^2 = 4kx\) and its latus rectum is \(\frac{8}{3} k^2\).

Solving for the Value of k

We are given that the area of the region is 24 square units. So, we can set up the equation:

$$\frac{8}{3} k^2 = 24$$

Now, we solve for \(k\):

Multiply both sides by 3:

$$8k^2 = 24 \times 3$$

$$8k^2 = 72$$

Divide both sides by 8:

$$k^2 = \frac{72}{8}$$

$$k^2 = 9$$

Take the square root of both sides:

$$k = \pm \sqrt{9}$$

$$k = \pm 3$$

The problem statement specifies that \(k > 0\). Therefore, we must choose the positive value of \(k\).

$$k = 3$$

Verification

If \(k=3\), the parabola is \(y^2 = 4(3)x = 12x\). The latus rectum is \(x=3\). The area bounded by \(y^2=12x\) and \(x=3\) is \(\frac{8}{3} k^2 = \frac{8}{3} (3)^2 = \frac{8}{3} \times 9 = 8 \times 3 = 24\) square units, which matches the given information.

The value of \(k\) is 3.

Revision Table: Key Concepts for Parabola y² = 4ax

Additional Information on Parabola Area Calculation

The formula for the area bounded by the parabola \(y^2 = 4ax\) and its latus rectum \(x=a\) is a standard result that can be derived using integration. The integration approach involves setting up the definite integral \(\int y \, dx\) over the appropriate limits. For \(y^2=4ax\), we integrate from \(x=0\) (vertex) to \(x=a\) (latus rectum). Since the parabola is symmetric about the x-axis, the total area is twice the area above the x-axis.

Area \(= 2 \int_{0}^{a} \sqrt{4ax} \, dx = 2 \int_{0}^{a} 2\sqrt{a}\sqrt{x} \, dx = 4\sqrt{a} \int_{0}^{a} x^{1/2} \, dx\). Evaluating this integral gives \(4\sqrt{a} \left[ \frac{2}{3} x^{3/2} \right]_{0}^{a} = 4\sqrt{a} \left(\frac{2}{3} a^{3/2}\right) = \frac{8}{3} a^{1/2} a^{3/2} = \frac{8}{3} a^2\).

This formula \(\frac{8}{3} a^2\) is applicable for parabolas of the form \(y^2 = 4ax\) or \(y^2 = -4ax\) bounding the area with their respective latus rectums \(x=a\) or \(x=-a\), where 'a' is the positive distance from the vertex to the focus. For the given problem, \(a=k\), hence the area is \(\frac{8}{3} k^2\).