The Tangency Condition for an Ellipse: A Core Concept

This problem leverages a fundamental property of ellipses: the condition for a straight line to be tangent to an ellipse. Understanding this condition is crucial for solving problems involving tangents to conic sections.

For an ellipse with the standard equation: $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$ And a straight line with the equation in slope-intercept form: $y = mx + c$ The line is tangent to the ellipse if and only if the following condition holds true: $c^2 = A^2m^2 + B^2$ Here, $A^2$ is the denominator of the $x^2$ term and $B^2$ is the denominator of the $y^2$ term in the ellipse equation. $m$ is the slope of the line, and $c$ is its y-intercept.

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Step 1: Convert the Tangent Line to Slope-Intercept Form ($y = mx + c$)

The given equation of the tangent line is $3x + 4y = 12\sqrt{2}$. To apply the tangency condition, we first need to express this line in the standard slope-intercept form, $y = mx + c$. This allows us to directly identify the slope ($m$) and the y-intercept ($c$).

Let's rearrange the equation: $4y = -3x + 12\sqrt{2}$ Divide both sides by 4: $y = -\frac{3}{4}x + \frac{12\sqrt{2}}{4}$ $y = -\frac{3}{4}x + 3\sqrt{2}$

Now, comparing this with $y = mx + c$: We identify the slope $m = -\frac{3}{4}$ and the y-intercept $c = 3\sqrt{2}$.

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Step 2: Identify Parameters of the Ellipse

The given equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{9} = 1$. Comparing this with the general form $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$:

We identify the parameters $A^2$ and $B^2$ for the tangency condition:

• $A^2 = a^2$ (the denominator of the $x^2$ term)
• $B^2 = 9$ (the denominator of the $y^2$ term)

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Step 3: Apply the Tangency Condition to Find the Unknown Parameter 'a'

Now we have all the necessary components:

• $m = -\frac{3}{4}$
• $c = 3\sqrt{2}$
• $A^2 = a^2$
• $B^2 = 9$

Substitute these values into the tangency condition $c^2 = A^2m^2 + B^2$: $(3\sqrt{2})^2 = (a^2)\left(-\frac{3}{4}\right)^2 + 9$

Let's calculate the squared terms: $(3\sqrt{2})^2 = 3^2 \times (\sqrt{2})^2 = 9 \times 2 = 18$ $\left(-\frac{3}{4}\right)^2 = \frac{(-3)^2}{4^2} = \frac{9}{16}$

Substitute these back into the equation: $18 = a^2 \left(\frac{9}{16}\right) + 9$

Now, solve for $a^2$: $18 - 9 = a^2 \left(\frac{9}{16}\right)$ $9 = a^2 \left(\frac{9}{16}\right)$ To isolate $a^2$, multiply both sides by $\frac{16}{9}$: $a^2 = 9 \times \frac{16}{9}$ $a^2 = 16$

So, we have found that $a^2 = 16$.

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Step 4: Determine the Semi-Major and Semi-Minor Axes of the Ellipse

The ellipse equation is $\frac{x^2}{16} + \frac{y^2}{9} = 1$. For an ellipse, the larger denominator under $x^2$ or $y^2$ determines the square of the semi-major axis. Here, we have $16$ and $9$. Since $16 > 9$:

• The semi-major axis squared is $A_{major}^2 = 16$, so the semi-major axis is $A_{major} = \sqrt{16} = 4$.
• The semi-minor axis squared is $B_{minor}^2 = 9$, so the semi-minor axis is $B_{minor} = \sqrt{9} = 3$.

Since $A_{major}^2$ is under the $x^2$ term, the major axis of the ellipse lies along the x-axis.

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Step 5: Calculate the Eccentricity (e) of the Ellipse

The eccentricity $e$ is a measure of how "stretched" an ellipse is. It's related to the semi-major axis ($A_{major}$) and semi-minor axis ($B_{minor}$) by the formula: $B_{minor}^2 = A_{major}^2 (1 - e^2)$ Alternatively, we can write it as: $e^2 = 1 - \frac{B_{minor}^2}{A_{major}^2}$

Using the values $A_{major}^2 = 16$ and $B_{minor}^2 = 9$: $e^2 = 1 - \frac{9}{16}$ To subtract, find a common denominator: $e^2 = \frac{16}{16} - \frac{9}{16}$ $e^2 = \frac{7}{16}$ Now, take the square root to find $e$: $e = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{\sqrt{16}} = \frac{\sqrt{7}}{4}$ The eccentricity of the ellipse is $e = \frac{\sqrt{7}}{4}$.

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Step 6: Calculate the Distance Between the Foci

The foci of an ellipse are located at a distance of $A_{major}e$ from the center along the major axis. Therefore, the distance between the two foci is $2 \times A_{major}e$.

Using $A_{major} = 4$ and $e = \frac{\sqrt{7}}{4}$: Distance between foci $= 2 \times A_{major} \times e$ $= 2 \times 4 \times \frac{\sqrt{7}}{4}$ $= 2 \times \sqrt{7}$ $= 2\sqrt{7}$

Therefore, the distance between the foci of the ellipse is $2\sqrt{7}$.

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Tips for Success and Common Mistakes to Avoid:

1. Standard Form is Key: Always convert the line equation to $y = mx + c$ and identify $A^2, B^2$ correctly from the ellipse equation before applying the tangency condition.
2. Distinguish 'a' in problem from 'semi-major axis': In this problem, the variable 'a' was given as part of the ellipse equation $\frac{x^2}{a^2}$. This 'a' is not necessarily the semi-major axis. After finding $a^2=16$, you must compare it with the other denominator ($9$) to correctly identify which is $A_{major}^2$ and $B_{minor}^2$.
3. Eccentricity Formula: Remember the correct formula for eccentricity: $B_{minor}^2 = A_{major}^2(1-e^2)$. Ensure you use the larger value for $A_{major}^2$. If the major axis were along the y-axis, the formula would be $A_{minor}^2 = B_{major}^2(1-e^2)$. A simpler way is $e^2 = 1 - \frac{\text{minor axis}^2}{\text{major axis}^2}$.
4. Foci Distance: The distance between foci is always $2 \times (\text{semi-major axis}) \times e$.

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Summary and Key Takeaway:

This problem demonstrates a classic application of the tangency condition for an ellipse. By systematically converting the line to slope-intercept form, identifying ellipse parameters, applying the tangency formula to find the unknown, and then using the ellipse properties (semi-major/minor axes and eccentricity), we can determine the distance between the foci. The key takeaway is the power of the tangency condition $c^2 = A^2m^2 + B^2$ as a bridge between the algebraic representation of a line and an ellipse.

The final answer is \boxed{\text{2\sqrt 7}}.