(A) 1/2√(5/3) (B) 1/2√(11/3) (C) √(5/6) (D) 1/3√(11/3)

Q: Wait, why is the length of the minor axis $2b$?

In the standard form of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the vertices are at $(\pm a, 0)$ and $(0, \pm b)$. The major axis connects the vertices on the longer axis, and the minor axis connects the vertices on the shorter axis. Since the minor axis is along the y-axis, its endpoints are $(0, b)$ and $(0, -b)$, making the total length $2b$.

Q: Where does the condition $c^2 = a^2m^2 + b^2$ come from?

This is a classic result! If you substitute $y = mx + c$ into the ellipse equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, you get a quadratic equation in $x$. For the line to be a tangent, this quadratic must have exactly one solution, meaning its discriminant must be zero. Solving $D = 0$ leads directly to $c^2 = a^2m^2 + b^2$.

Q: How do I know if the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ or $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$?

The problem explicitly states the minor axis is along the y-axis. This means the ellipse is 'wider' along the x-axis. Therefore, the denominator under $x^2$ must be the larger value ($a^2$) and the denominator under $y^2$ must be the smaller value ($b^2$). Thus, the form is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$.

Question

The length of minor axis (along y-axis) of an ellipse of the standard form is $4/\sqrt{3}$. If this ellipse touches the line $x + 6y = 8$, then its eccentricity is :

Options:

(A) $\frac{1}{2}\sqrt{\frac{5}{3}}$

(B) $\frac{1}{2}\sqrt{\frac{11}{3}}$

(C) $\sqrt{\frac{5}{6}}$

(D) $\frac{1}{3}\sqrt{\frac{11}{3}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a canvas, drawing a perfect, symmetrical oval. This is the ellipse, a shape that has fascinated mathematicians from Kepler to the present day.

In our problem, we are given a standard ellipse, defined by the elegant equation:

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

The problem states that the minor axis lies along the $y$-axis. This implies the ellipse is stretched horizontally, meaning $a > b$.

The length of this minor axis is given as $\frac{4}{\sqrt{3}}$. Since the total length of the minor axis is $2b$, we deduce:

$$ b = \frac{2}{\sqrt{3}} \implies b^2 = \frac{4}{3} $$

This is our first anchor point in the sea of variables.

### The Tangent Dance

Now, consider a line passing through our coordinate plane: $x + 6y = 8$. This line is a tangent, touching the ellipse at exactly one point.

To work with this, we rearrange the equation into the slope-intercept form $y = mx + c$:

$$ 6y = -x + 8 \implies y = -\frac{1}{6}x + \frac{4}{3} $$

Here, the slope $m$ is $-\frac{1}{6}$ and the $y$-intercept $c$ is $\frac{4}{3}$. This transformation is vital for applying the condition of tangency.

### The Condition of Tangency

In the world of JEE Advanced, the condition for a line $y = mx + c$ to be tangent to an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by the formula:

$$ c^2 = a^2m^2 + b^2 $$

We have all the pieces of the puzzle: $c = \frac{4}{3}$, $m = -\frac{1}{6}$, and $b^2 = \frac{4}{3}$. Substituting these into our equation:

$$ \left( \frac{4}{3} ight)^2 = a^2 \left( -\frac{1}{6} ight)^2 + \frac{4}{3} $$

### Solving for the Ellipse

Performing the algebra with care, we square the terms:

$$ \frac{16}{9} = a^2 \left( \frac{1}{36} ight) + \frac{4}{3} $$

To isolate $a^2$, we subtract $\frac{4}{3}$ (or $\frac{12}{9}$) from both sides:

$$ \frac{16}{9} - \frac{12}{9} = \frac{a^2}{36} \implies \frac{4}{9} = \frac{a^2}{36} $$

Multiplying both sides by $36$, we find:

$$ a^2 = 36 	imes \frac{4}{9} = 16 $$

### The Final Step: Eccentricity

The eccentricity $e$ of an ellipse is defined by the formula:

$$ e = \sqrt{1 - \frac{b^2}{a^2}} $$

Substituting $a^2 = 16$ and $b^2 = \frac{4}{3}$, we get:

$$ e = \sqrt{1 - \frac{4/3}{16}} = \sqrt{1 - \frac{4}{48}} = \sqrt{1 - \frac{1}{12}} $$

Simplifying the expression, we arrive at the final result:

$$ e = \sqrt{\frac{11}{12}} = \frac{\sqrt{11}}{2\sqrt{3}} = \mathbf{\frac{1}{2}\sqrt{\frac{11}{3}}} $$

Animated Solution for Mathematics - Conic Sections: The length of minor axis (along y-axis) of an ellipse of the standard form is $4/ 3$ . If this ellipse touches the line $x + 6 y = 8$ , then its eccentricity is :

Select Answer:

Visualized Solution

Standard Ellipse and Tangent

Length of Minor Axis

Finding $b^{2}$

Slope-Intercept Form

Extracting $m$ and $c$

Condition for Tangency

Substituting Knowns

Simplifying Squares

Finding $a^{2}$

Eccentricity Formula

Calculating Eccentricity

Final Result

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent Dance

The Condition of Tangency

Solving for the Ellipse

The Final Step

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