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

In coordinate geometry, for any conic section (circle, parabola, ellipse, or hyperbola) represented by the general equation $S=0$, the equation of a chord whose midpoint is a given point $(x_1, y_1)$ can be found using a powerful and efficient formula: $T = S_1$ This formula is a cornerstone for solving problems involving chords and conics, providing a direct method without needing to find the endpoints of the chord or calculate slopes.

Let's break down the terms in this formula:

• $S=0$: This represents the general equation of the conic section, arranged such that all terms are on one side, equating to zero. For example, for an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we write $S: \frac{x^2}{a^2} + \frac{y^2}{b^2} - 1 = 0$.
• $S_1$: This is the value obtained by substituting the coordinates of the given midpoint $(x_1, y_1)$ directly into the equation of the conic $S=0$. That is, $S_1 = S(x_1, y_1)$. The sign of $S_1$ provides geometric insight:
  • If $S_1 < 0$, the point $(x_1, y_1)$ lies inside the conic. This is a necessary condition for it to be the midpoint of a real chord.
  • If $S_1 > 0$, the point $(x_1, y_1)$ lies outside the conic.
  • If $S_1 = 0$, the point $(x_1, y_1)$ lies on the conic. In this special case, the "chord" becomes a tangent at that point.
• $T$: This is an expression derived from the conic's equation by making specific substitutions based on the coordinates $(x_1, y_1)$. The general substitution rules are:
  • $x^2 \to xx_1$
  • $y^2 \to yy_1$
  • $x \to \frac{x+x_1}{2}$
  • $y \to \frac{y+y_1}{2}$
  • $xy \to \frac{xy_1+yx_1}{2}$ Essentially, $T$ represents the expression for the tangent to the conic at $(x_1, y_1)$ if $(x_1, y_1)$ were on the conic, or more generally, it is the equation of the polar of the point $(x_1, y_1)$ with respect to the conic. When equated to $S_1$, it yields the equation of the chord whose midpoint is $(x_1, y_1)$.

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Problem Statement and Given Information

We are asked to find the equation of the chord of the given ellipse, whose midpoint is specified.

• The equation of the ellipse: $\frac{x^2}{25}+\frac{y^2}{16}=1$
• The coordinates of the midpoint of the chord: $(x_1, y_1) = (3, 1)$

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Step-by-Step Solution

Step 1: Express the Ellipse Equation in the Standard Form ($S=0$) The first step is to rearrange the given ellipse equation so that all terms are on one side, equating to zero. This helps in clearly identifying the expression for $S$ for subsequent calculations. $S: \frac{x^2}{25}+\frac{y^2}{16}-1=0$ Why this step? This standardization is crucial for correctly applying the $T=S_1$ formula. It ensures that the constant term is properly included in both $S_1$ and $T$ calculations.

Step 2: Calculate $S_1$ Next, we substitute the coordinates of the given midpoint $(x_1, y_1) = (3, 1)$ into the $S$ expression derived in Step 1. This will give us the value of $S_1$. $S_1 = \frac{(3)^2}{25}+\frac{(1)^2}{16}-1$ $S_1 = \frac{9}{25}+\frac{1}{16}-1$ To simplify this expression, we need to find a common denominator for the fractions. The least common multiple (LCM) of 25 and 16 is $25 \times 16 = 400$. $S_1 = \frac{9 \times 16}{25 \times 16}+\frac{1 \times 25}{16 \times 25}-1$ $S_1 = \frac{144}{400}+\frac{25}{400}-1$ Combine the fractions: $S_1 = \frac{144+25}{400}-1$ $S_1 = \frac{169}{400}-1$ Now, express $1$ as $\frac{400}{400}$ and combine: $S_1 = \frac{169-400}{400}$ $S_1 = -\frac{231}{400}$ Why this step? $S_1$ is a crucial component of the $T=S_1$ formula. The negative value of $S_1$ confirms that the midpoint $(3,1)$ lies inside the ellipse, which is a necessary and expected condition for it to be the midpoint of a real chord. If $S_1$ were positive, the point would be outside the ellipse, and no real chord could have it as a midpoint.

Step 3: Calculate $T$ Now, we form the expression for $T$ by applying the substitution rules to the ellipse equation $S=0$, using the midpoint coordinates $(x_1, y_1) = (3, 1)$. The ellipse equation is $\frac{x^2}{25}+\frac{y^2}{16}-1=0$. According to the rules:

• Replace $x^2$ with $xx_1$
• Replace $y^2$ with $yy_1$
• Constant terms remain unchanged. Substituting $(x_1, y_1) = (3, 1)$: $T: \frac{x(3)}{25}+\frac{y(1)}{16}-1$ $T: \frac{3x}{25}+\frac{y}{16}-1$ Why this step? $T$ is the other crucial component of the $T=S_1$ formula. This process effectively linearizes the non-linear equation of the ellipse into a linear expression. When this linear expression is equated to $S_1$, it yields the equation of the desired chord.

Step 4: Apply the Chord Formula $T=S_1$ With $T$ and $S_1$ calculated, we can now directly apply the formula: $T = S_1$ $\frac{3x}{25}+\frac{y}{16}-1 = -\frac{231}{400}$ Why this step? This is the direct application of the fundamental formula, which immediately gives us the equation of the chord in an unsimplified form.

Step 5: Simplify the Equation to Standard Linear Form To obtain the equation in a cleaner, standard linear form ($Ax+By=C$), we will perform algebraic manipulations. First, move the constant term from the left side to the right side of the equation: $\frac{3x}{25}+\frac{y}{16} = 1 - \frac{231}{400}$ Combine the constant terms on the right side: $\frac{3x}{25}+\frac{y}{16} = \frac{400}{400} - \frac{231}{400}$ $\frac{3x}{25}+\frac{y}{16} = \frac{400-231}{400}$ $\frac{3x}{25}+\frac{y}{16} = \frac{169}{400}$ To eliminate the denominators and simplify the equation further, we multiply the entire equation by the LCM of 25 and 16, which is 400: $400 \times \left(\frac{3x}{25}\right) + 400 \times \left(\frac{y}{16}\right) = 400 \times \left(\frac{169}{400}\right)$ Perform the multiplications: $16(3x) + 25y = 169$ $48x + 25y = 169$ Why this step? This final simplification presents the equation of the chord in a standard, easily understandable linear form, making it straightforward to compare with the given options and identify the correct answer.

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Matching with Options

The derived equation of the chord is $48x + 25y = 169$. Comparing this with the given options: (A) $5 x+16 y=31$ (B) $48 x+25 y=169$ (C) $4 x+122 y=134$ (D) $25 x+101 y=176$

The calculated equation matches option (B).

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Important Tips and Common Pitfalls

• Master the $T=S_1$ Formula: This formula is incredibly versatile and applicable to all conic sections. Memorizing it will save significant time in exams compared to alternative methods (like assuming chord endpoints $(x_2, y_2)$ and $(x_3, y_3)$, using the midpoint formula, and then finding the slope).
• Correct Substitutions for $T$: Pay close attention to the specific substitution rules for each term ($x^2 \to xx_1$, $y^2 \to yy_1$, etc.). A common mistake is to substitute $x_1$ for $x$ or $y_1$ for $y$ in the $T$ expression, which would incorrectly make $T$ a constant instead of a linear equation. Remember that $x$ and $y$ in $T$ are the variables of the chord equation.
• Algebraic Precision: Be meticulous with fraction arithmetic, finding LCMs, and simplifying expressions. Small calculation errors can lead to a completely different final equation.
• Standard Form $S=0$: Always ensure the conic equation is in the $S=0$ form before calculating $S_1$ and $T$. Missing a constant term or having terms on the wrong side can lead to errors.
• Geometric Check: A quick mental check on the sign of $S_1$ ($S_1 < 0$ for a midpoint inside the ellipse) can help confirm if your midpoint is valid for a real chord.

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Conclusion and Key Takeaway

This problem demonstrates the efficiency and elegance of using the $T=S_1$ formula to find the equation of a chord of a conic section with a given midpoint. By systematically applying the definitions of $S$, $S_1$, and $T$, we can directly derive the linear equation of the chord. This method is a fundamental tool in coordinate geometry for JEE aspirants.