Properties of Ellipse with Common Tangents to Circle and Parabola

To find the properties of the given ellipse and the bounded area, we proceed step-by-step.

Step 1: Find the Common Tangents to the Parabola and the Circle

The given equations are:

1. Circle: $x^2 + y^2 = \frac{1}{2}$, centered at $O(0,0)$ with radius $r = \frac{1}{\sqrt{2}}$.
2. Parabola: $y^2 = 4x$, where $a = 1$.

Any tangent line to the parabola $y^2 = 4x$ with slope $m$ is given by: $y = mx + \frac{1}{m} \implies m^2 x - my + 1 = 0$

Since this line is also tangent to the circle $x^2 + y^2 = \frac{1}{2}$, the perpendicular distance from the origin $(0,0)$ to the line must be equal to the radius $r$: $\frac{|1|}{\sqrt{m^4 + (-m)^2}} = \frac{1}{\sqrt{2}}$ $\sqrt{m^4 + m^2} = \sqrt{2}$

Squaring both sides gives: $m^4 + m^2 - 2 = 0$ $(m^2 + 2)(m^2 - 1) = 0$

Since $m \in \mathbb{R}$, we have $m^2 = 1 \implies m = \pm 1$.

Thus, the equations of the common tangents are:

• For $m = 1$: $x - y + 1 = 0$
• For $m = -1$: $x + y + 1 = 0$

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Step 2: Determine the Intersection Point $Q$ and Ellipse Parameters

Solving the system of equations for the two tangent lines: $x - y + 1 = 0$ $x + y + 1 = 0$

Adding both equations gives $2x + 2 = 0 \implies x = -1$, which yields $y = 0$. So, the intersection point is $Q(-1, 0)$.

• The semi-major axis is given as $OQ$. The length of $OQ$ is: $a = \text{distance}(O, Q) = \sqrt{(-1 - 0)^2 + (0 - 0)^2} = 1$ Since $Q$ lies on the x-axis, the major axis of the ellipse lies along the x-axis.

• The length of the minor axis is given as $\sqrt{2}$: $2b = \sqrt{2} \implies b = \frac{1}{\sqrt{2}}$

Thus, the equation of the ellipse is: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \implies x^2 + 2y^2 = 1$

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Step 3: Evaluate Options A and B

1. Eccentricity ($e$): $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1/2}{1}} = \frac{1}{\sqrt{2}}$

2. Length of the Latus Rectum ($LR$): $LR = \frac{2b^2}{a} = \frac{2 \left(\frac{1}{2}\right)}{1} = 1$

Therefore, Statement (A) is TRUE, and Statement (B) is FALSE.

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Step 4: Evaluate Options C and D (Bounded Area)

The region bounded by the ellipse between $x = \frac{1}{\sqrt{2}}$ and $x = 1$ is symmetric about the x-axis. Expressing $y$ in terms of $x$: $y = \pm \frac{1}{\sqrt{2}}\sqrt{1 - x^2}$

The total area $A$ of the region is given by: $A = 2 \int_{1/\sqrt{2}}^{1} y \, dx = 2 \int_{1/\sqrt{2}}^{1} \frac{1}{\sqrt{2}}\sqrt{1 - x^2} \, dx = \sqrt{2} \int_{1/\sqrt{2}}^{1} \sqrt{1 - x^2} \, dx$

Using the standard integration formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin\left(\frac{x}{a}\right)$:

$\int_{1/\sqrt{2}}^{1} \sqrt{1 - x^2} \, dx = \left[ \frac{x}{2}\sqrt{1 - x^2} + \frac{1}{2}\arcsin(x) \right]_{1/\sqrt{2}}^{1}$

Evaluating at the upper limit $x = 1$: $\frac{1}{2}(0) + \frac{1}{2}\arcsin(1) = \frac{\pi}{4}$

Evaluating at the lower limit $x = \frac{1}{\sqrt{2}}$: $\frac{1}{2\sqrt{2}}\sqrt{1 - \frac{1}{2}} + \frac{1}{2}\arcsin\left(\frac{1}{\sqrt{2}}\right) = \frac{1}{4} + \frac{\pi}{8}$

Subtracting the lower limit value from the upper limit value: $\int_{1/\sqrt{2}}^{1} \sqrt{1 - x^2} \, dx = \frac{\pi}{4} - \left(\frac{1}{4} + \frac{\pi}{8}\right) = \frac{\pi - 2}{8}$

Multiplying by $\sqrt{2}$: $A = \sqrt{2} \left(\frac{\pi - 2}{8}\right) = \frac{1}{4\sqrt{2}}(\pi - 2)$

Therefore, Statement (C) is TRUE, and Statement (D) is FALSE.

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Conclusion

The correct statements are A and C.