The length of the chord of the ellipse x²/4 + y²/2 = 1 whose mid-point is (1, 1/2) is:
(1) 5/3 √15
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Q: Wait, why does $T = S_1$ actually work? Is it just a magic formula?

It is not magic, but a beautiful consequence of the geometry of conics! $T$ represents the tangent expression at a point, and $S_1$ is the power of the point. When you set $T = S_1$, you are essentially defining the locus of points that satisfy the condition of being the midpoint of a chord. It is a shortcut derived from the general equation of a conic, and mastering it will save you precious minutes in the exam hall.

Q: Why can't I just find the intersection points by solving the quadratic equation fully?

You absolutely can! But in the JEE Advanced arena, time is your most precious currency. Solving for $x$ and $y$ individually often leads to messy radicals and a higher probability of arithmetic errors. Using the difference of roots $|x_1 - x_2| = \frac{\sqrt{D}}{|a|}$ allows you to bypass the calculation of the individual coordinates entirely. It is cleaner, faster, and much more robust.

Q: Why do we use the formula $L = |x_1 - x_2| \sqrt{1 + m^2}$? Where does it come from?

Think of the chord as the hypotenuse of a right-angled triangle. The horizontal leg is the difference in x-coordinates, $|x_1 - x_2|$, and the vertical leg is the difference in y-coordinates, $|y_2 - y_1|$. Since the slope $m = \frac{y_2 - y_1}{x_2 - x_1}$, we know that $|y_2 - y_1| = |m| \cdot |x_2 - x_1|$. By the Pythagorean theorem, the length is $\sqrt{(x_2-x_1)^2 + (m(x_2-x_1))^2}$, which simplifies perfectly to $|x_1 - x_2| \sqrt{1 + m^2}$.

Question

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:

Options:

(A) $\frac{5}{3}\sqrt{15}$

(B) $\frac{1}{3}\sqrt{15}$

(C) $\frac{2}{3}\sqrt{15}$

(D) $\sqrt{15}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are uncovering the hidden architecture of an ellipse.

We are given the equation:

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

Because the denominator under $x^2$ is larger than the one under $y^2$, we know this ellipse is stretched horizontally, a graceful oval resting on the Cartesian plane. We are tasked with finding the length of a specific chord—a line segment that cuts through this ellipse, anchored perfectly at its midpoint, $P(1, \frac{1}{2})$.

### The Magic of $T = S_1$

When you encounter a problem involving a 'chord with a given midpoint', your mind should immediately race to one of the most elegant tools in the JEE toolkit: the $T = S_1$ theorem. This allows us to bypass the tedious process of finding the slope of the chord or defining it with arbitrary variables.

To use this, we define $T$ as the tangent expression at the point $(x_1, y_1)$ and $S_1$ as the value of the ellipse equation at that same point. For our ellipse, we replace $x^2$ with $x \cdot x_1$ and $y^2$ with $y \cdot y_1$. Substituting our midpoint $P(1, \frac{1}{2})$, we get:

$$T = \frac{x(1)}{4} + \frac{y(1/2)}{2} = \frac{x}{4} + \frac{y}{4}$$

Next, we calculate $S_1$ by simply plugging the midpoint into the ellipse equation:

$$S_1 = \frac{1^2}{4} + \frac{(1/2)^2}{2} = \frac{1}{4} + \frac{1}{8} = \frac{3}{8}$$

Equating $T = S_1$, we arrive at $\frac{x}{4} + \frac{y}{4} = \frac{3}{8}$. Multiplying by $8$ gives us the beautiful, simple linear equation: $2x + 2y = 3$, or $y = \frac{3}{2} - x$. This line is the unique chord that bisects the ellipse at our chosen point.

### The Intersection: Where Algebra Meets Geometry

Now that we have the equation of the chord, we need to find where it pierces the ellipse. We substitute $y = \frac{3}{2} - x$ into the original ellipse equation, which we multiply by $4$ to get $x^2 + 2y^2 = 4$.

Substituting our expression for $y$:

$$x^2 + 2\left(\frac{3}{2} - x\right)^2 = 4$$

Expanding this requires care: $x^2 + 2(\frac{9}{4} - 3x + x^2) = 4$. Distributing the $2$, we get $x^2 + \frac{9}{2} - 6x + 2x^2 = 4$.

Combining like terms, we arrive at $3x^2 - 6x + \frac{1}{2} = 0$. Multiplying by $2$ to clear the fraction, we get the quadratic:

$$6x^2 - 12x + 1 = 0$$

### The Final Leap: The Difference of Roots

The length of the chord is given by the formula $L = |x_1 - x_2| \sqrt{1 + m^2}$. We know the slope $m$ of our line $y = -x + \frac{3}{2}$ is $-1$. Thus, $\sqrt{1 + m^2} = \sqrt{1 + (-1)^2} = \sqrt{2}$.

For the difference of roots $|x_1 - x_2|$, we use the formula $\frac{\sqrt{D}}{|a|}$, where $D = b^2 - 4ac$. For our quadratic $6x^2 - 12x + 1 = 0$:

$$D = (-12)^2 - 4(6)(1) = 144 - 24 = 120$$

So, $|x_1 - x_2| = \frac{\sqrt{120}}{6} = \frac{2\sqrt{30}}{6} = \frac{\sqrt{30}}{3}$. Finally, multiplying by our slope factor:

$$L = \frac{\sqrt{30}}{3} \cdot \sqrt{2} = \frac{\sqrt{60}}{3} = \frac{2\sqrt{15}}{3}$$

The length of the chord is **$\frac{2\sqrt{15}}{3}$**. You have successfully navigated the geometry, the algebra, and the final calculation.

Animated Solution for Mathematics - Conic Sections: 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:

Select Answer:

Visualized Solution

Visualizing the Ellipse

Locating the Midpoint

Drawing the Chord

The $T = S_{1}$ Concept

Calculating $T$

Calculating $S_{1}$

Forming the Chord Equation

Finding Intersection Points

Expanding the Equation

Simplifying the Quadratic

Difference of Roots

Simplifying Difference of Roots

Calculating Chord Length

Final Result

The Sigma Insight: Chord in terms of Midpoint

Analyzing the Setup

The Magic of $T = S_{1}$

The Intersection

The Final Leap

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The length of the chord of the ellipse 4x2​+2y2​=1 whose mid-point is (1,21​) is: