Area inside the parabola y² = 4ax between the lines x = a & x = 4a is equal to

• a) 4a²
• b) 8a²
• c) 28/3 a²
• d) none of these

To find the area inside the parabola \(y^2 = 4ax\) between the lines \(x = a\) and \(x = 4a\), we can use integration. The parabola opens to the right, and we can express \(y\) in terms of \(x\) as \(y = \sqrt{4ax}\) and \(y = -\sqrt{4ax}\).

Setting Up the Integral

The area \(A\) can be calculated using the formula:

A = ∫[a to 4a] (upper curve - lower curve) dx

In this case, the upper curve is \(y = \sqrt{4ax}\) and the lower curve is \(y = -\sqrt{4ax}\). Therefore, the area becomes:

Calculating the Area

A = ∫[a to 4a] (√(4ax) - (-√(4ax))) dx

A = ∫[a to 4a] 2√(4ax) dx

A = 2∫[a to 4a] 2√(4a)√x dx

A = 4√(4a) ∫[a to 4a] √x dx

Evaluating the Integral

The integral of √x is:

∫√x dx = (2/3)x^(3/2)

Now, we evaluate it from \(a\) to \(4a\):

A = 4√(4a) * [(2/3)(4a)^(3/2) - (2/3)(a)^(3/2)]

Final Calculation

Calculating the values:

• (4a)^(3/2) = 8a√a
• (a)^(3/2) = a√a

Substituting these back, we get:

A = 4√(4a) * (2/3)(8a√a - 2a√a)

A = 4√(4a) * (2/3)(6a√a) = 16a²√(4a)/3

After simplifying, we find that the area is:

A = (32/3)a²

Conclusion

Thus, the area inside the parabola between the given lines is equal to:

None of the provided options are correct.