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

Answer: (A)

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

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.