Understanding the Core Concept: Tangent to an Ellipse at a Point

The foundation of this problem lies in the ability to write the equation of a tangent line to an ellipse when the point of tangency is known. For a standard ellipse defined by the equation ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$ if a point $(x_1, y_1)$ lies on the ellipse, then the equation of the tangent line to the ellipse at this specific point $(x_1, y_1)$ is given by the formula: ${{x{x_1}} \over {{a^2}}} + {{y{y_1}} \over {{b^2}}} = 1$ This formula is derived using calculus (finding the derivative $dy/dx$ at $(x_1, y_1)$ to get the slope, and then using the point-slope form of a line) or through implicit differentiation. It's a fundamental result in coordinate geometry for ellipses and is crucial for many problems involving tangency.

Our goal is to use this formula, along with the given tangent line, to determine the parameters $a^2$ and $b^2$ of the ellipse, and then use them to calculate the length of the latus rectum.

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Step-by-Step Solution

Step 1: Formulate the equation of the tangent using the given point of tangency.

We are given the general equation of the ellipse as ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$. We are also provided with the point of tangency, $(x_1, y_1) = \left( {3, - {9 \over 2}} \right)$. Our first step is to substitute these coordinates into the general tangent equation formula: ${{x(3)} \over {{a^2}}} + {{y\left( { - {9 \over 2}} \right)} \over {{b^2}}} = 1$ Simplifying this expression, we get: ${{3x} \over {{a^2}}} - {{9y} \over {2{b^2}}} = 1 \quad \text{(Equation 1)}$ Explanation: This step directly applies the tangent formula. By substituting the coordinates of the point of tangency, we obtain an equation for the tangent line that involves the unknown parameters $a^2$ and $b^2$ of the ellipse. This equation represents the specific tangent line at the given point, but its coefficients are expressed in terms of $a^2$ and $b^2$.

Step 2: Rewrite the given tangent equation in a comparable standard form.

We are given that the tangent line is $x - 2y = 12$. To effectively compare this equation with Equation 1 (which has 1 on the right-hand side), it is essential to rewrite the given tangent equation in a similar form. This normalization makes the direct comparison of coefficients straightforward and reduces the chance of errors related to proportionality constants. We achieve this by dividing the entire equation by 12: ${{x} \over {12}} - {{2y} \over {12}} = {{12} \over {12}}$ ${{x} \over {12}} - {{y} \over 6} = 1 \quad \text{(Equation 2)}$ Explanation: Two linear equations represent the exact same line if and only if their corresponding coefficients are proportional. By manipulating both tangent equations such that their constant terms are identical (in this case, both equal to 1), we ensure that the coefficients of $x$ and $y$ must also be exactly equal, not just proportional. This simplifies the comparison process significantly.

Step 3: Equate corresponding coefficients to solve for $a^2$ and $b^2$.

Now we have two equations that both represent the same tangent line: From Equation 1: ${{3x} \over {{a^2}}} - {{9y} \over {2{b^2}}} = 1$ From Equation 2: ${{x} \over {12}} - {{y} \over 6} = 1$

Since these two equations describe the identical line, and we have normalized their constant terms to 1, their respective coefficients for $x$ must be equal, and similarly, their respective coefficients for $y$ must be equal.

Equating coefficients of $x$: ${3 \over {{a^2}}} = {1 \over {12}}$ To solve for $a^2$, we cross-multiply: $3 \times 12 = 1 \times {a^2}$ ${a^2} = 36$

Equating coefficients of $y$: ${{ - 9} \over {2{b^2}}} = {{ - 1} \over 6}$ First, we can multiply both sides by -1 to eliminate the negative signs, making the calculation clearer: ${9 \over {2{b^2}}} = {1 \over 6}$ Now, cross-multiply to solve for $b^2$: $9 \times 6 = 1 \times 2{b^2}$ $54 = 2{b^2}$ ${b^2} = {{54} \over 2}$ ${b^2} = 27$ Explanation: This step is the core of finding the ellipse's parameters. By recognizing that both Equation 1 and Equation 2 represent the same line and are normalized, we can directly equate the coefficients of $x$ and $y$. This provides two simple algebraic equations, which we then solve to find the specific values of $a^2$ and $b^2$ that define the unique ellipse.

Step 4: Calculate the length of the latus rectum of the ellipse.

The length of the latus rectum is a fundamental characteristic that describes the "width" of the ellipse at its foci. For an ellipse of the form ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$, the formula for the length of the latus rectum depends on whether the major axis is along the x-axis or the y-axis, which is determined by whether $a > b$ or $b > a$.

From our calculations, we have $a^2 = 36$ and $b^2 = 27$. Taking the positive square root for lengths (as $a$ and $b$ represent semi-axes lengths): $a = \sqrt{36} = 6$ $b = \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$ Now, let's compare $a$ and $b$: $a = 6$ $b = 3\sqrt{3} \approx 3 \times 1.732 = 5.196$ Since $a = 6$ and $b \approx 5.196$, we clearly have $a > b$. This indicates that the major axis of the ellipse lies along the x-axis.

For this configuration ($a > b$), the length of the latus rectum is given by the formula: ${\text{Length of Latus Rectum}} = {{2{b^2}} \over a}$ Now, substitute the values of $a=6$ and $b^2=27$ into the formula: ${\text{Length of Latus Rectum}} = {{2 \times 27} \over 6}$ ${\text{Length of Latus Rectum}} = {{54} \over 6}$ ${\text{Length of Latus Rectum}} = 9$ Explanation: The latus rectum is a chord passing through a focus and perpendicular to the major axis. Its length is a key descriptor of the ellipse's shape. We've correctly identified the values of $a$ and $b$ and confirmed that $a > b$, which means the major axis is horizontal. Therefore, the formula $2b^2/a$ is the correct one to use for the length of the latus rectum in this case. Plugging in our derived values for $a$ and $b^2$ gives us the final answer.

The final answer is $\boxed{\text{9}}$.

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Important Tips for Success & Common Mistakes

• Verify the Point of Tangency: Always remember that the given point of tangency must lie on both the ellipse and the tangent line. While not strictly required to solve this particular problem (as it's given that it's the point of tangency), it's an excellent habit for verification. For instance, check if $(3, -9/2)$ satisfies the given tangent line equation $x - 2y = 12$: $3 - 2(-9/2) = 3 + 9 = 12$. This confirms the point is indeed on the given line. You could also (optionally) verify if this point lies on the ellipse $x^2/36 + y^2/27 = 1$: $3^2/36 + (-9/2)^2/27 = 9/36 + (81/4)/27 = 1/4 + 81/(4 \times 27) = 1/4 + 3/4 = 1$. This confirms the point is indeed on the ellipse we found.
• Standardize Equations for Comparison: When comparing two linear equations (say, $A_1x + B_1y = C_1$ and $A_2x + B_2y = C_2$) that represent the same line, it's safest to make their constant terms equal (e.g., both equal to 1, or both equal to 0, or simply use the proportionality $A_1/A_2 = B_1/B_2 = C_1/C_2$). This prevents errors if you directly equate coefficients without proper normalization.
• Correct Latus Rectum Formula: Pay close attention to the values of $a$ and $b$. For an ellipse $x^2/A^2 + y^2/B^2 = 1$:
  • If $A > B$, then $a=A$, $b=B$, the major axis is horizontal, and the latus rectum length is $2b^2/a = 2B^2/A$.
  • If $B > A$, then $a=B$, $b=A$, the major axis is vertical, and the latus rectum length is $2a^2/b = 2A^2/B$. Always calculate $a$ and $b$ first (from $A^2$ and $B^2$) and then apply the appropriate formula based on which value is larger. In this case, $a=6$ and $b=3\sqrt{3} \approx 5.19$, so $a>b$, and the formula $2b^2/a$ is correct.
• Sign Errors: Be careful with negative signs, especially when comparing coefficients of $y$ if one equation has a positive $y$ term and the other has a negative $y$ term. Ensure you carry the sign correctly.

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Summary and Key Takeaways

This problem is a classic example testing your understanding of fundamental concepts related to ellipses in coordinate geometry. The key steps involved are:

1. Tangent Equation Application: Correctly applying the formula for the tangent to an ellipse at a given point to express the tangent in terms of the ellipse's parameters ($a^2, b^2$).
2. Equation Normalization: Transforming the given linear equation of the tangent into a form directly comparable to the derived tangent equation.
3. Coefficient Comparison: Using the principle that two identical lines must have proportional (or equal, if normalized) coefficients to set up and solve equations for $a^2$ and $b^2$.
4. Latus Rectum Calculation: Determining the major and minor axes lengths ($a$ and $b$) and then applying the correct formula for the length of the latus rectum based on the orientation of the major axis.

Mastering these steps ensures you can confidently tackle similar problems involving tangents and properties of ellipses.