Question

The parabola \( y^2 = 4x \) divides the area of the circle \( x^2 + y^2 = 5 \) in two parts. The area of the smaller part is equal to:

• A. \( \frac{2}{3} + 5 \sin^{-1}\left( \frac{2}{\sqrt{5}} \right) \)
• B. \( \frac{1}{3} + 5 \sin^{-1}\left( \frac{2}{\sqrt{5}} \right) \)
• C. \( \frac{1}{3} + \sqrt{5} \sin^{-1}\left( \frac{2}{\sqrt{5}} \right) \)
• D. \( \frac{2}{3} + \sqrt{5} \sin^{-1}\left( \frac{2}{\sqrt{5}} \right) \)

Answer 1

The Correct Option is A

Given the equations: 

\( y^2 = 4x \) and \( x^2 + y^2 = 5 \)

∴ The area of the shaded region as shown in the figure will be:

\( A_1 = \int_{0}^{1} \sqrt{4x} \, dx + \int_{1}^{\sqrt{5}} \sqrt{5 - x^2} \, dx \)

Now, evaluating the integrals:

\( A_1 = \frac{4}{3} \left[ x^{3/2} \right]_{0}^{1} + \left[ \frac{x}{2} \sqrt{5 - x^2} + \frac{5}{2} \sin^{-1} \left( \frac{x}{\sqrt{5}} \right) \right]_{1}^{\sqrt{5}} \)

Substituting the limits:

\( A_1 = \frac{1}{3} + \frac{5\pi}{4} - \frac{5}{2} \sin^{-1} \left( \frac{1}{\sqrt{5}} \right) \)

∴ Required area \( = 2A_1 \)

\( = \frac{2}{3} + \frac{5\pi}{2} - 5 \sin^{-1} \left( \frac{1}{\sqrt{5}} \right) \)

Since \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \), we can write:

\( = \frac{2}{3} + 5 \cos^{-1} \left( \frac{1}{\sqrt{5}} \right) \)

Or equivalently,

\( = \frac{2}{3} + 5 \sin^{-1} \left( \frac{2}{\sqrt{5}} \right) \)

Answer 2

The parabola \( y^2 = 4x \) intersects the circle \( x^2 + y^2 = 5 \) at two points. To find these points of intersection, substitute \( x = \frac{y^2}{4} \) into \( x^2 + y^2 = 5 \):

\[ \left( \frac{y^2}{4} \right)^2 + y^2 = 5. \]

Simplify:

\[ \frac{y^4}{16} + y^2 = 5. \]

Multiply through by 16:

\[ y^4 + 16y^2 = 80. \]

Let \( z = y^2 \), so:

\[ z^2 + 16z - 80 = 0. \]

Solve this quadratic equation:

\[ z = \frac{-16 \pm \sqrt{16^2 - 4(1)(-80)}}{2(1)} = \frac{-16 \pm \sqrt{256 + 320}}{2} = \frac{-16 \pm \sqrt{576}}{2}. \] \[ z = \frac{-16 \pm 24}{2}. \]

This gives \( z = 4 \) or \( z = -20 \) (discarded as \( z = y^2 \geq 0 \)).

Thus, \( y^2 = 4 \), so \( y = \pm 2 \). The points of intersection are \( (1, 2) \) and \( (1, -2) \).

The region to the left of the parabola is bounded by the circle \( x^2 + y^2 = 5 \), and we calculate the area of this smaller region.

Area Calculation

The area of the circle segment is:

\[ A_{\text{segment}} = R^2 \sin^{-1} \left( \frac{d}{R} \right) - \frac{d}{2} \sqrt{R^2 - d^2}, \]

where \( R = \sqrt{5} \) is the radius of the circle and \( d = 2 \) is the distance from the circle center to the chord.

Substitute the values:

\[ A_{\text{segment}} = 5 \sin^{-1} \left( \frac{2}{\sqrt{5}} \right) - \frac{2}{2} \sqrt{5 - 4}. \]

The area of the parabola sector is:

\[ A_{\text{parabola}} = \frac{2}{3}. \]

Thus, the smaller part’s area is:

\[ A_{\text{smaller}} = \frac{2}{3} + 5 \sin^{-1} \left( \frac{2}{\sqrt{5}} \right). \]
Thus the correct answer is Option 1.