Solution

This problem asks us to find the sum of coefficients $\alpha$ and $\beta$ from the equation of a chord of an ellipse, given its midpoint. The most efficient way to solve this is by using the fundamental formula for the equation of a chord when its midpoint is known.

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1. Key Concept: Equation of a Chord with a Given Midpoint ($T=S_1$)

For any general conic section represented by the equation $S=0$, if we know the midpoint of a chord as $P(x_1, y_1)$, the equation of that chord is given by the formula: $T = S_1$ Let's break down what each term means for a general second-degree equation $S \equiv Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$:

• $S$: This is the original equation of the conic section, set to zero.
• $T$: This term is obtained by performing specific substitutions in $S$.
  • Replace $x^2$ with $xx_1$
  • Replace $y^2$ with $yy_1$
  • Replace $xy$ with $\frac{xy_1+yx_1}{2}$
  • Replace $x$ with $\frac{x+x_1}{2}$
  • Replace $y$ with $\frac{y+y_1}{2}$
  • Constant terms remain unchanged.
• $S_1$: This is the value obtained by substituting the coordinates of the midpoint $(x_1, y_1)$ into the original equation $S=0$. So, $S_1 = Ax_1^2 + Bx_1y_1 + Cy_1^2 + Dx_1 + Ey_1 + F$.

Applying to an Ellipse:

For an ellipse given by the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we can write $S \equiv \frac{x^2}{a^2} + \frac{y^2}{b^2} - 1 = 0$.

• Calculating $T$: Using the substitution rules for $x^2$ and $y^2$: $T = \frac{xx_1}{a^2} + \frac{yy_1}{b^2} - 1$

• Calculating $S_1$: Substituting $(x_1, y_1)$ into $S$: $S_1 = \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} - 1$

Now, equating $T=S_1$: $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} - 1 = \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} - 1$ Notice that the constant term '$-1$' appears on both sides of the equation. This term cancels out, leading to a simplified and highly useful form for the equation of the chord of an ellipse whose midpoint is $(x_1, y_1)$: $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2}$ This is the formula we will use to solve the problem.

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2. Step-by-Step Solution

Step 1: Identify the Given Information

First, we extract all the necessary information from the problem statement:

• The equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{4} = 1$.
  • By comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we identify $a^2 = 9$ and $b^2 = 4$.
• The midpoint of the chord is given as $\left(\frac{5}{2}, \frac{1}{2}\right)$.
  • So, we have $x_1 = \frac{5}{2}$ and $y_1 = \frac{1}{2}$.
• The target form of the chord equation is $\alpha x + \beta y = 109$. We need to find $\alpha + \beta$.

Step 2: Apply the $T=S_1$ Formula for the Ellipse

Now, we substitute the values of $x_1, y_1, a^2,$ and $b^2$ into the simplified $T=S_1$ formula for an ellipse: $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2}$ Substituting the values: $\frac{x\left(\frac{5}{2}\right)}{9} + \frac{y\left(\frac{1}{2}\right)}{4} = \frac{\left(\frac{5}{2}\right)^2}{9} + \frac{\left(\frac{1}{2}\right)^2}{4}$ This step is crucial as it directly sets up the equation of the chord based on the given midpoint and ellipse parameters.

Step 3: Simplify the Left-Hand Side (LHS) of the Equation

Let's simplify the terms involving $x$ and $y$: $\frac{x \cdot 5}{2 \cdot 9} + \frac{y \cdot 1}{2 \cdot 4}$ $= \frac{5x}{18} + \frac{y}{8}$ This simplifies the structure of the linear equation in $x$ and $y$.

Step 4: Simplify the Right-Hand Side (RHS) of the Equation

Next, we simplify the constant terms on the RHS: $\frac{\left(\frac{5}{2}\right)^2}{9} + \frac{\left(\frac{1}{2}\right)^2}{4}$ Calculate the squares: $= \frac{\frac{25}{4}}{9} + \frac{\frac{1}{4}}{4}$ Convert complex fractions to simpler ones: $= \frac{25}{4 \times 9} + \frac{1}{4 \times 4}$ $= \frac{25}{36} + \frac{1}{16}$ To add these fractions, we need to find their Least Common Multiple (LCM) of the denominators 36 and 16.

• Prime factorization of $36 = 2^2 \times 3^2$
• Prime factorization of $16 = 2^4$
• LCM$(36, 16) = 2^4 \times 3^2 = 16 \times 9 = 144$.

Now, we rewrite the fractions with the common denominator 144: $= \frac{25 \times 4}{36 \times 4} + \frac{1 \times 9}{16 \times 9}$ $= \frac{100}{144} + \frac{9}{144}$ $= \frac{100+9}{144}$ $= \frac{109}{144}$ This step ensures the constant part of the chord equation is correctly calculated.

Step 5: Combine and Clear Denominators

Now, we put the simplified LHS and RHS back together: $\frac{5x}{18} + \frac{y}{8} = \frac{109}{144}$ To make the equation look like $\alpha x + \beta y = \text{constant}$ and eliminate fractions, we multiply the entire equation by the LCM of all denominators (18, 8, and 144).

• LCM$(18, 8, 144) = 144$ (as 18 and 8 are factors of 144).

Multiply every term by 144: $144 \left(\frac{5x}{18}\right) + 144 \left(\frac{y}{8}\right) = 144 \left(\frac{109}{144}\right)$ Perform the multiplications and cancellations: $(8 \times 5x) + (18 \times y) = 109$ $40x + 18y = 109$ This gives us the final simplified equation of the chord in the desired linear form.

Step 6: Compare with the Given Chord Equation

The problem states that the equation of the chord is $\alpha x + \beta y = 109$. We derived the equation of the chord as $40x + 18y = 109$.

By comparing the coefficients of $x$ and $y$ and the constant term, we can directly identify:

• $\alpha = 40$
• $\beta = 18$ This comparison is valid because the constant term on the RHS (109) matches in both equations. If they were different, we would need to scale our derived equation to match the constant term before comparing $\alpha$ and $\beta$.

Step 7: Calculate $\alpha + \beta$

Finally, we calculate the required sum: $\alpha + \beta = 40 + 18$ $\alpha + \beta = 58$

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3. Tips and Common Mistakes

• Correct $T=S_1$ Application: Always ensure you use the correct form of $T$ and $S_1$ for the given conic. For an ellipse, the constant term canceling out is a helpful simplification, but remember the general formula if the conic's equation is not centered at the origin or contains $x$ or $y$ terms.
• Arithmetic Precision: The most common errors in such problems arise from arithmetic mistakes, especially when dealing with fractions and finding LCMs. Double-check all calculations, particularly for squaring fractions and adding them.
• LCM Calculation: Be systematic in finding the LCM of denominators. Prime factorization is a reliable method.
• Coefficient Comparison: Ensure the constant term on the RHS of your derived equation matches the given chord equation before directly comparing $\alpha$ and $\beta$. If they differ, scale your equation appropriately. For example, if you got $20x + 9y = \frac{109}{2}$, you'd multiply by 2 to get $40x + 18y = 109$.

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4. Summary and Key Takeaway

This problem is a direct application of the $T=S_1$ formula, a powerful tool in coordinate geometry for finding equations of tangents, chords with given midpoints, and polar equations for conic sections. The process involves:

1. Identifying the parameters of the conic ($a^2, b^2$) and the midpoint coordinates ($x_1, y_1$).
2. Substituting these values into the specific $T=S_1$ formula for an ellipse.
3. Carefully simplifying the resulting equation by performing fractional arithmetic and clearing denominators.
4. Comparing the final simplified equation with the given form to determine the unknown coefficients.

Mastering the $T=S_1$ formula and careful algebraic manipulation are key to solving such problems efficiently and accurately in competitive exams like JEE.

The final answer is $\boxed{\text{58}}$.