JEE Main 2025 (January)

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Animated Solution for Mathematics - Conic Sections: If $α x + β y = 109$ is the equation of the chord of the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1.$ whose mid point is $(\frac{5}{2}, \frac{1}{2})$ then $α + β$ is equal to:

Q: Wait, why is the formula $T = S_1$ and not $T = 0$?

Great question! $T = 0$ is the equation of the tangent at a point *on* the curve. But our midpoint is *inside* the ellipse, not on it. $T = S_1$ is the generalized form that works for any chord with a given midpoint.

Select Answer:

Progressive Hints

Hint 1

The Spark: Visualize the ellipse. A chord is defined by its endpoints, but when you are given the midpoint, you are essentially given the 'center of gravity' of that chord. Don't try to find the endpoints; that is a long, winding road. Instead, focus on the property that the chord is uniquely determined by its midpoint.

Hint 2

The Arsenal: In the world of conic sections, there is a legendary tool for this exact scenario: the formula

T = S_{1}

. This is not just a formula; it is a geometric identity that relates the tangent-like expression

T

at the midpoint to the power of the point

S_{1}

Hint 3

The Execution: Start by defining your ellipse as

S = \frac{x ^{2}}{9} + \frac{y ^{2}}{4} - 1 = 0

. Now, calculate

T

by replacing

x^{2}

with

x x_{1}

and

y^{2}

with

y y_{1}

. Then, calculate

S_{1}

by plugging the midpoint

(x_{1}, y_{1}) = (\frac{5}{2}, \frac{1}{2})

into the ellipse equation. Finally, equate

T = S_{1}

and simplify to match the form

α x + β y = 109

Visualized Solution

Visualizing the Ellipse and Midpoint

Given Ellipse: $\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$
Midpoint of the chord: $M (x_{1}, y_{1}) = (\frac{5}{2}, \frac{1}{2})$

The Chord Equation

A chord passes through $M$ .
Given equation of this chord: $α x + β y = 109$
Target: Find the value of $α + β$ .

The Master Formula: $T = S_{1}$

The equation of a chord with a given midpoint $(x_{1}, y_{1})$ is:
$T = S_{1}$
$T$ : Tangent expression at $(x_{1}, y_{1})$
$S_{1}$ : Power of the point $(x_{1}, y_{1})$

Constructing the $T$ Expression

To find $T$ , use the transformations:
$x^{2} \to x x_{1}$
$y^{2} \to y y_{1}$
For the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{4} - 1 = 0$ :
$T = \frac{x x _{1}}{9} + \frac{y y _{1}}{4} - 1$

Substituting Midpoint into $T$

Substitute $(x_{1}, y_{1}) = (\frac{5}{2}, \frac{1}{2})$ :
$T = \frac{x ( \frac{5}{2} )}{9} + \frac{y ( \frac{1}{2} )}{4} - 1$
$T = \frac{5 x}{18} + \frac{y}{8} - 1$

Constructing the $S_{1}$ Expression

To find $S_{1}$ , substitute $(x_{1}, y_{1})$ directly into the ellipse expression.
$S_{1} = \frac{x _{1}^{2}}{9} + \frac{y _{1}^{2}}{4} - 1$
Substitute $(x_{1}, y_{1}) = (\frac{5}{2}, \frac{1}{2})$ :
$S_{1} = \frac{( \frac{5}{2} ) ^{2}}{9} + \frac{( \frac{1}{2} ) ^{2}}{4} - 1$

Simplifying $S_{1}$

Square the numerators:
$S_{1} = \frac{\frac{25}{4}}{9} + \frac{\frac{1}{4}}{4} - 1$
$S_{1} = \frac{25}{36} + \frac{1}{16} - 1$

Calculating the LCM for $S_{1}$

Find the common denominator for 36 and 16, which is 144.
$S_{1} = \frac{25 \times 4}{144} + \frac{1 \times 9}{144} - 1$
$S_{1} = \frac{100 + 9}{144} - 1$
$S_{1} = \frac{109}{144} - 1$

Equating $T$ and $S_{1}$

Set $T = S_{1}$ :
$\frac{5 x}{18} + \frac{y}{8} - 1 = \frac{109}{144} - 1$
Cancel $- 1$ from both sides:
$\frac{5 x}{18} + \frac{y}{8} = \frac{109}{144}$

Clearing the Denominators

Multiply the entire equation by $144$ :
$144 \cdot (\frac{5 x}{18}) + 144 \cdot (\frac{y}{8}) = 144 \cdot (\frac{109}{144})$
$8 (5 x) + 18 (y) = 109$
$40 x + 18 y = 109$

Comparing with Given Equation

Derived Equation: $40 x + 18 y = 109$
Given Equation: $α x + β y = 109$
By comparing coefficients:
$α = 40$
$β = 18$

The Final Answer

Calculate the required sum:
$α + β = 40 + 18$
$α + β = 58$
Correct Option: 58

00:00 / 00:00

The Sigma Insight: Chord in terms of Midpoint

Solution Diagram

Analyzing the Setup

We are given the ellipse defined by the equation:

\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1

We seek the equation of a chord that has its midpoint at

M (\frac{5}{2}, \frac{1}{2})

. The final equation must be expressed in the form

α x + β y = 109

The Master Equation

In coordinate geometry, the equation of a chord of a conic section with a given midpoint

(x_{1}, y_{1})

is elegantly provided by the formula:

T = S_{1}

Here,

T

represents the expression obtained by replacing

x^{2}

with

x x_{1}

and

y^{2}

with

y y_{1}

in the ellipse equation, while

S_{1}

is the value of the ellipse equation evaluated at the point

(x_{1}, y_{1})

Constructing the Components

First, we construct

T

using the midpoint

M (\frac{5}{2}, \frac{1}{2})

T = \frac{x ( \frac{5}{2} )}{9} + \frac{y ( \frac{1}{2} )}{4} - 1 = \frac{5 x}{18} + \frac{y}{8} - 1

Next, we calculate

S_{1}

by substituting the coordinates of

M

into the ellipse equation:

S_{1} = \frac{( \frac{5}{2} ) ^{2}}{9} + \frac{( \frac{1}{2} ) ^{2}}{4} - 1 = \frac{25/4}{9} + \frac{1/4}{4} - 1

Simplifying the fractions, we obtain:

S_{1} = \frac{25}{36} + \frac{1}{16} - 1 = \frac{100 + 9}{144} - 1 = \frac{109}{144} - 1

Final Calculation

Equating

T = S_{1}

, we have:

\frac{5 x}{18} + \frac{y}{8} - 1 = \frac{109}{144} - 1

The constant terms cancel out, leaving:

\frac{5 x}{18} + \frac{y}{8} = \frac{109}{144}

To transform this into the required form

α x + β y = 109

, we multiply the entire equation by

144

144 (\frac{5 x}{18}) + 144 (\frac{y}{8}) = 109

40 x + 18 y = 109

Comparing this to

α x + β y = 109

, we identify

α = 40

and

β = 18

. Therefore, the sum is $α + β = 58$ .

Common Traps & FAQs

Why can't I just find the slope of the chord using the midpoint?

That is a common trap! You don't know the slope of the chord initially. The midpoint *determines* the slope. If you assume a slope

m

, you'll end up with a system of equations that is much harder to solve than simply using the

T = S_{1}

identity.

What if I tried to find the intersection points of the chord with the ellipse? Is that faster?

That would be algebraic suicide. Finding the intersection points requires solving a quadratic equation, which is unnecessary. The

T = S_{1}

formula is the elegant shortcut designed specifically for this purpose.

Wait, why is the formula

T = S_{1}

and not

T = 0

Great question!

T = 0

is the equation of the tangent at a point *on* the curve. But our midpoint is *inside* the ellipse, not on it.

T = S_{1}

is the generalized form that works for any chord with a given midpoint.

Animated Solution for Mathematics - Conic Sections: If $α x + β y = 109$ is the equation of the chord of the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1.$ whose mid point is $(\frac{5}{2}, \frac{1}{2})$ then $α + β$ is equal to:

Select Answer:

Visualized Solution

Visualizing the Ellipse and Midpoint

The Chord Equation

The Master Formula: $T = S_{1}$

Constructing the $T$ Expression

Substituting Midpoint into $T$

Constructing the $S_{1}$ Expression

Simplifying $S_{1}$

Calculating the LCM for $S_{1}$

Equating $T$ and $S_{1}$

Clearing the Denominators

Comparing with Given Equation

The Final Answer

The Sigma Insight: Chord in terms of Midpoint

Analyzing the Setup

The Master Equation

Constructing the Components

Final Calculation

Similar Questions

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The equation of the chord, of the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$ whose mid-point is $(3, 1)$ is:

The length of the chord of the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{2} = 1$ whose mid-point is $(1, \frac{1}{2})$ is:

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If a chord, which is not a tangent, of the parabola $y^{2} = 16 x$ has the equation $2 x + y = p$ , and midpoint $(h, k)$ , then which of the following is(are) possible value(s) of $p, h$ and $k$ ?

The locus of the mid points of the chords of the hyperbola $x^{2} - y^{2} = 4$ , which touch the parabola $y^{2} = 8 x$ , is :

Let $P Q$ be a chord of the parabola $y^{2} = 12 x$ and the midpoint of $P Q$ be at $(4, 1)$ . Then, which of the following point lies on the line passing through the points $P$ and $Q$ ?

If two distinct chords, drawn from the point $(p, q)$ on the circle $x^{2} + y^{2} = p x + q y$ (where $pq \neq = 0$ ) are bisected by the x-axis, then

Select Answer:

Visualized Solution

Visualizing the Ellipse and Midpoint

The Chord Equation

The Master Formula: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=S1​

Constructing the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T Expression

Substituting Midpoint into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

Constructing the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​ Expression

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​

Calculating the LCM for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​

Equating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​

Clearing the Denominators

Comparing with Given Equation

The Final Answer

The Sigma Insight: Chord in terms of Midpoint

Analyzing the Setup

The Master Equation

Constructing the Components

Final Calculation

Similar Questions

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The Master Formula: $T = S_{1}$

Constructing the $T$ Expression

Substituting Midpoint into $T$

Constructing the $S_{1}$ Expression

Simplifying $S_{1}$

Calculating the LCM for $S_{1}$

Equating $T$ and $S_{1}$

If the midpoint of a chord of the ellipseis $\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ is $(2, 4/3)$ and the length of the chord is $\frac{2 α}{3}$ then a is:

The equation of the chord, of the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$ whose mid-point is $(3, 1)$ is:

The length of the chord of the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{2} = 1$ whose mid-point is $(1, \frac{1}{2})$ is:

Let $P (α, β)$ be a point on the parabola $y^{2} = 4 x$ . If $P$ also lies on the chord of the parabola $x^{2} = 8 y$ whose mid point is $(1, \frac{5}{4})$ . Then $(α - 28) (β - 8)$ is equal to ______.

If a chord, which is not a tangent, of the parabola $y^{2} = 16 x$ has the equation $2 x + y = p$ , and midpoint $(h, k)$ , then which of the following is(are) possible value(s) of $p, h$ and $k$ ?

The locus of the mid points of the chords of the hyperbola $x^{2} - y^{2} = 4$ , which touch the parabola $y^{2} = 8 x$ , is :

Let $P Q$ be a chord of the parabola $y^{2} = 12 x$ and the midpoint of $P Q$ be at $(4, 1)$ . Then, which of the following point lies on the line passing through the points $P$ and $Q$ ?

If two distinct chords, drawn from the point $(p, q)$ on the circle $x^{2} + y^{2} = p x + q y$ (where $pq \neq = 0$ ) are bisected by the x-axis, then