If the midpoint of a chord of the ellipseis x²/9 + y²/4 = 1 is (√2, 4/3) and the length of the chord

Q: Wait, why do we use $T = S_1$ instead of the slope-point form $y - y_1 = m(x - x_1)$?

You absolutely could use the slope-point form, but it would require you to first find the slope $m$ by considering the condition that the line intersects the ellipse at two points whose midpoint is $(x_1, y_1)$. The formula $T = S_1$ is a direct, elegant shortcut derived specifically for this scenario. It saves you from unnecessary algebraic complexity and potential errors.

Q: Is there a faster way to find the length of the chord without finding the intersection points?

Yes, there is a formula for the length of a chord with a given midpoint $(x_1, y_1)$ for an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, which is $L = \frac{2ab \sqrt{(a^2y_1^2 + b^2x_1^2 - a^2b^2)(a^2 + b^2)}}{a^2y_1^2 + b^2x_1^2}$. However, for JEE Advanced, deriving it through the intersection points is often safer and less prone to memory-based errors.

Q: Why did the constant term $36$ cancel out when I substituted the line into the ellipse?

That is a beautiful feature of the geometry! When you substitute the equation of a chord (which is a line) into the equation of the ellipse, the constant terms on both sides of the equation must cancel out if the line is indeed a chord passing through the ellipse. If they didn't cancel, it would imply the line doesn't intersect the ellipse in the way we expect. It's a great way to check if your algebra is on the right track.

Question

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

Options:

(A) 20

(B) 22

(C) 18

(D) 26

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The ellipse is defined by the equation:

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

We are given a chord with its midpoint $M$ located at $(\sqrt{2}, \frac{4}{3})$. In coordinate geometry, the equation of a chord with a given midpoint $(x_1, y_1)$ is determined by the **$T = S_1$** theorem.

### The Magic of $T = S_1$

The expression $T$ is derived by replacing $x^2$ with $x \cdot x_1$ and $y^2$ with $y \cdot y_1$. Substituting the midpoint $(\sqrt{2}, \frac{4}{3})$, we obtain:

$$ T = \frac{x \cdot \sqrt{2}}{9} + \frac{y \cdot \frac{4}{3}}{4} = \frac{\sqrt{2}x}{9} + \frac{y}{3} $$

Next, we calculate $S_1$ by evaluating the ellipse equation at the midpoint:

$$ S_1 = \frac{(\sqrt{2})^2}{9} + \frac{(\frac{4}{3})^2}{4} = \frac{2}{9} + \frac{16/9}{4} = \frac{2}{9} + \frac{4}{9} = \frac{6}{9} = \frac{2}{3} $$

Equating $T = S_1$, we get $\frac{\sqrt{2}x}{9} + \frac{y}{3} = \frac{2}{3}$. Multiplying the entire equation by $9$ yields the linear equation of the chord:

$$ \sqrt{2}x + 3y = 6 $$

### The Algebra of Intersection

To find the intersection points, we rearrange the chord equation to $y = 2 - \frac{\sqrt{2}}{3}x$ and substitute it into the ellipse equation $4x^2 + 9y^2 = 36$.

Expanding the term $9(2 - \frac{\sqrt{2}}{3}x)^2$ results in:

$$ 9\left(4 - \frac{4\sqrt{2}}{3}x + \frac{2}{9}x^2\right) = 36 - 12\sqrt{2}x + 2x^2 $$

Adding the $4x^2$ from the original ellipse equation, the constant $36$ cancels out, leaving the quadratic:

$$ 6x^2 - 12\sqrt{2}x = 0 $$

Factoring this expression gives $6x(x - 2\sqrt{2}) = 0$, which provides the $x$-coordinates $x_1 = 0$ and $x_2 = 2\sqrt{2}$.

### The Final Stretch

Using the chord equation, we find the corresponding $y$-coordinates. For $x_1 = 0$, we find $y_1 = 2$. For $x_2 = 2\sqrt{2}$, we find $y_2 = 2 - \frac{\sqrt{2}}{3}(2\sqrt{2}) = \frac{2}{3}$.

The endpoints of the chord are $A(0, 2)$ and $B(2\sqrt{2}, \frac{2}{3})$. We calculate the length $L$ using the distance formula:

$$ L = \sqrt{(2\sqrt{2} - 0)^2 + \left(\frac{2}{3} - 2\right)^2} = \sqrt{8 + \left(-\frac{4}{3}\right)^2} = \sqrt{8 + \frac{16}{9}} $$

$$ L = \sqrt{\frac{72 + 16}{9}} = \sqrt{\frac{88}{9}} = \frac{2\sqrt{22}}{3} $$

Comparing this result to the form $\frac{2\sqrt{\alpha}}{3}$, we conclude that **$\alpha = 22$**.

Animated Solution for Mathematics - Conic Sections: 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:

Select Answer:

Visualized Solution

Visualize the Ellipse and Midpoint

The $T = S_{1}$ Concept

Calculate Expression $T$

Calculate Constant $S_{1}$

Formulate the Chord Equation

Substitute into Ellipse Equation

Expand and Simplify

Solve for $x$ -coordinates

Find $y$ -coordinates

Calculate Chord Length

Final Comparison and Result

The Sigma Insight: Chord in terms of Midpoint

Analyzing the Setup

The Magic of $T = S_{1}$

The Algebra of Intersection

The Final Stretch

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Visualized Solution

Visualize the Ellipse and Midpoint

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

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

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Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-coordinates