This problem is a classic application of analytical geometry, specifically involving ellipses and tangents, combined with trigonometric optimization. Our goal is to find the minimum possible area of a triangle formed by a tangent line to a given ellipse and the coordinate axes.

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Key Concepts and Formulas (Prerequisites)

To tackle this problem, we need to be proficient with the following fundamental concepts:

1. Standard Equation of an Ellipse: An ellipse centered at the origin has the standard form ${{{x^2}} \over {{A^2}}} + {{{y^2}} \over {{B^2}}} = 1$. Here, $A$ and $B$ represent the lengths of the semi-major and semi-minor axes along the x and y directions, respectively.
2. Parametric Coordinates on an Ellipse: Any point on the ellipse ${{{x^2}} \over {{A^2}}} + {{{y^2}} \over {{B^2}}} = 1$ can be conveniently represented in parametric form as $(A\cos\theta, B\sin\theta)$. This representation simplifies many geometric calculations.
3. Equation of a Tangent to an Ellipse (Parametric Form): The equation of the tangent line to the ellipse ${{{x^2}} \over {{A^2}}} + {{{y^2}} \over {{B^2}}} = 1$ at the point $(A\cos\theta, B\sin\theta)$ is given by ${{x\cos \theta } \over A} + {{y\sin \theta } \over B} = 1$. This form is particularly useful for finding intercepts.
4. Area of a Triangle formed by a Line and Coordinate Axes: If a straight line intersects the x-axis at $(X_0, 0)$ and the y-axis at $(0, Y_0)$, the area of the triangle formed by this line and the coordinate axes (with the origin as one vertex) is ${1 \over 2} |X_0 Y_0|$. For typical minimum area problems, we consider the tangent in the first quadrant, where $X_0 > 0$ and $Y_0 > 0$, simplifying the area to ${1 \over 2} X_0 Y_0$.
5. Trigonometric Double-Angle Identity: The identity $\sin(2\theta) = 2\sin\theta\cos\theta$ is crucial for simplifying expressions involving products of sine and cosine functions.
6. Range of the Sine Function: The value of $\sin x$ is always bounded between -1 and 1, i.e., $-1 \le \sin x \le 1$. Consequently, its maximum positive value is 1. This property is fundamental for optimization (finding minimum or maximum values).

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Detailed Step-by-Step Solution

Let's break down the problem into logical steps, explaining the reasoning behind each action.

Step 1: Understanding the Ellipse and its Tangent Equation

Our first task is to correctly identify the parameters of the given ellipse and then formulate the general equation of a tangent to it.

• Given Ellipse Equation: The problem provides the ellipse equation as ${{{x^2}} \over {{b^2}}} + {{{y^2}} \over {4{a^2}}} = 1$

• Why this step? To apply the standard formulas for tangents and other properties, we must first correctly identify the semi-axis lengths, often denoted as $A$ and $B$, from the given equation. These values will be substituted into the general formulas.

• Action: We compare the given equation with the standard form of an ellipse centered at the origin, ${{{x^2}} \over {{A^2}}} + {{{y^2}} \over {{B^2}}} = 1$. From this comparison, we deduce: ${A^2} = {b^2} \implies A = b$ (The semi-axis length along the x-axis) ${B^2} = {4{a^2}} \implies B = 2a$ (The semi-axis length along the y-axis)

  Now, we write the equation of the tangent to this specific ellipse.

• Why parametric form? The parametric form of the tangent equation ${{x\cos \theta } \over A} + {{y\sin \theta } \over B} = 1$ is particularly advantageous here because it directly provides expressions for the x and y intercepts in terms of $\cos\theta$ and $\sin\theta$. These expressions are easy to manipulate for area calculation and subsequent minimization. Using the slope-intercept form ($y=mx \pm \sqrt{A^2m^2+B^2}$) would be more cumbersome for finding intercepts and then minimizing.

• Action: Substitute $A=b$ and $B=2a$ into the general tangent formula: ${{x\cos \theta } \over b} + {{y\sin \theta } \over {2a}} = 1$ This is the equation of the tangent line to the given ellipse at the point $(b\cos\theta, 2a\sin\theta)$.

Step 2: Finding the Intercepts with the Coordinate Axes

The triangle whose area we need to minimize is formed by this tangent line and the x and y axes. To calculate its area, we must first find where this tangent line intersects the coordinate axes.

• Why this step? The x-intercept and y-intercept will serve as the base and height (or legs) of the right-angled triangle formed with the origin. These lengths are essential for calculating the triangle's area.

• Action for x-intercept: To find the point where the tangent line crosses the x-axis, we set $y=0$ in the tangent equation: ${{x\cos \theta } \over b} + {{0\sin \theta } \over {2a}} = 1$ ${{x\cos \theta } \over b} = 1$ $x = {b \over {\cos \theta}}$ So, the x-intercept is $X_0 = {b \over {\cos \theta}}$.

• Action for y-intercept: Similarly, to find the point where the tangent line crosses the y-axis, we set $x=0$ in the tangent equation: ${{0\cos \theta } \over b} + {{y\sin \theta } \over {2a}} = 1$ ${{y\sin \theta } \over {2a}} = 1$ $y = {{2a} \over {\sin \theta}}$ So, the y-intercept is $Y_0 = {{2a} \over {\sin \theta}}$.

• Important Note: For the minimum area, the tangent line will typically be in the first quadrant, meaning the point of tangency $(b\cos\theta, 2a\sin\theta)$ is also in the first quadrant. This implies $\cos\theta > 0$ and $\sin\theta > 0$, which means $\theta \in (0, \pi/2)$. Consequently, both $X_0$ and $Y_0$ will be positive, allowing us to drop the absolute value signs in the area formula.

Step 3: Calculating the Area of the Triangle

Now that we have the x and y intercepts, we can calculate the area of the right-angled triangle formed by the tangent line and the coordinate axes.

• Why this step? We need an explicit expression for the area in terms of $\theta$ that we can then minimize.

• Action: Using the formula for the area of a triangle formed by a line and the coordinate axes: $Area (\Delta OAB) = {1 \over 2} |X_0 Y_0|$ Since $X_0$ and $Y_0$ are positive (as explained above for minimum area), we have: $Area = {1 \over 2} X_0 Y_0$ Substitute the intercept expressions we found: $Area = {1 \over 2} \times \left( {{b \over {\cos \theta }}} \right) \times \left( {{{2a} \over {\sin \theta }}} \right)$ $Area = {{2ab} \over {2\cos \theta \sin \theta}}$

Step 4: Simplifying the Area Expression using Trigonometric Identities

The current area expression contains a product of $\cos\theta$ and $\sin\theta$ in the denominator. This can be significantly simplified using a standard trigonometric identity.

• Why this step? Simplifying the expression is crucial because it allows us to reduce it to a single trigonometric function. This transformation makes it much easier to find the minimum value by leveraging the known range of that function.

• Action: We recall the double-angle identity: $\sin(2\theta) = 2\sin\theta\cos\theta$. Substitute this identity into our area expression: $Area = {{2ab} \over {\sin(2\theta)}}$

Step 5: Minimizing the Area

With the area now expressed in a simplified form, we can use the properties of the sine function to find its minimum value.

• Why this step? To find the minimum value of the area, we need to understand how the expression $Area = {{2ab} \over {\sin(2\theta)}}$ behaves as $\theta$ changes.

• Action: We want to minimize $Area = {{2ab} \over {\sin(2\theta)}}$. Since $2ab$ is a positive constant, to minimize the entire fraction, we must maximize its denominator, $\sin(2\theta)$. We know that the maximum value of the sine function, $\sin(x)$, is 1. Therefore, the maximum value of $\sin(2\theta)$ is 1. This maximum occurs when $2\theta = {\pi \over 2}$ (or any $2n\pi + \pi/2$), which implies $\theta = {\pi \over 4}$. This value of $\theta$ is indeed in the first quadrant $(0, \pi/2)$, validating our earlier assumption that the intercepts are positive.

  Substitute the maximum value of $\sin(2\theta) = 1$ into the area expression to find the minimum area: $Area_{min} = {{2ab} \over {1}}$ $Area_{min} = 2ab$

Step 6: Determining the Value of k

The problem states that the minimum area of the triangle is $kab$. We have just calculated this minimum area.

• Why this step? This is the final step to answer the specific question asked by comparing our derived minimum area with the given symbolic representation.

• Action: We are given that: $Minimum Area = kab$ From our calculations, we found that: $Minimum Area = 2ab$ By comparing these two expressions, we can equate them: $kab = 2ab$ Assuming $a$ and $b$ are non-zero (as they represent lengths of semi-axes and thus must be positive), we can divide both sides of the equation by $ab$: $k = 2$

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Important Tips and Common Pitfalls

• Correctly Identify A and B: A very common mistake is to automatically assume $A=a$ and $B=b$. Always carefully compare the given ellipse equation (here, ${{{x^2}} \over {{b^2}}} + {{{y^2}} \over {4{a^2}}} = 1$) with the standard form ${{{x^2}} \over {{A^2}}} + {{{y^2}} \over {{B^2}}} = 1$. In this problem, $A=b$ and $B=2a$. Misidentification will lead to incorrect intercepts and ultimately a wrong answer.
• Understanding Absolute Values for Area: While the general area formula is ${1 \over 2} |X_0 Y_0|$, for minimum area problems involving ellipses, the tangent point is invariably considered in the first quadrant ($\theta \in (0, \pi/2)$). This ensures that $\cos\theta$ and $\sin\theta$ are positive, making $X_0$ and $Y_0$ positive, which allows you to safely drop the absolute value