Let a line L pass through the point of intersection of the lines bx + 10y - 8 = 0 and 2x - 3y = 0, b

Q: Why use the family of lines instead of finding the intersection point first?

While you could solve the system of two linear equations to find the intersection point, it often leads to messy fractions or complex arithmetic. The 'Family of Lines' method allows you to represent all lines passing through that intersection point with a single parameter $\lambda$. It is an elegant, algebraic shortcut that keeps your calculations clean.

Q: Why is the distance from the origin to the line equal to the radius?

This is the fundamental geometric definition of a tangent to a circle. A line is tangent to a circle if and only if the shortest distance from the center of the circle to the line is exactly equal to the radius. Since our circle $x^2 + y^2 = \frac{16}{17}$ is centered at $(0, 0)$, the distance from $(0, 0)$ to our line $L$ must be $r = \frac{4}{\sqrt{17}}$.

Q: How do I know if the ellipse is horizontal or vertical?

The orientation of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ depends on the relative values of the denominators. If $a^2 > b^2$, the major axis is along the x-axis (horizontal). If $b^2 > a^2$, it is along the y-axis (vertical). In our case, we found $a^2 = 5$ and $b^2 = 2$. Since $5 > 2$, it is a horizontal ellipse, and we use the standard eccentricity formula $e = \sqrt{1 - \frac{b^2}{a^2}}$.

Question

Let a line $L$ pass through the point of intersection of the lines $bx + 10y - 8 = 0$ and $2x - 3y = 0, b \in \mathbb{R} - \{-\frac{4}{3}\}$. If the line $L$ also passes through the point $(1, 1)$ and touches the circle $17(x^2 + y^2) = 16$, then the eccentricity of the ellipse $\frac{x^2}{5} + \frac{y^2}{b^2} = 1$ is :

Options:

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

(B) $\sqrt{\frac{3}{5}}$

(C) $\frac{1}{\sqrt{5}}$

(D) $\sqrt{\frac{2}{5}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery! Today, we are not just solving a problem; we are orchestrating a symphony of coordinate geometry. We have a line $L$ dancing through the intersection of two other lines, kissing a circle, and finally defining the shape of an ellipse.

We are given two lines: $bx + 10y - 8 = 0$ and $2x - 3y = 0$. A novice might immediately try to solve these for $x$ and $y$. But you, as an elite student, know better. We use the **Family of Lines** theorem.

Any line passing through the intersection of $L_1$ and $L_2$ can be written as $L_1 + \lambda L_2 = 0$. Thus, our line $L$ is:

$$(bx + 10y - 8) + \lambda(2x - 3y) = 0$$

### The Power of the Family of Lines

This is powerful because we do not need to know the intersection point yet; we just need to know that $L$ passes through $(1, 1)$. Substituting $x=1$ and $y=1$ into our equation, we get:

$$(b + 10 - 8) + \lambda(2 - 3) = 0$$

This simplifies to $(b + 2) - \lambda = 0$, or simply $\lambda = b + 2$. With $\lambda$ in hand, we substitute it back to find the specific equation of $L$:

$$(bx + 10y - 8) + (b + 2)(2x - 3y) = 0$$

Expanding this and grouping the $x$ and $y$ terms:

$$(b + 2b + 4)x + (10 - 3b - 6)y - 8 = 0$$

This gives us the elegant line $L$:

$$(3b + 4)x + (4 - 3b)y - 8 = 0$$

### The Tangency Condition

Now, we turn to the circle $17(x^2 + y^2) = 16$. Dividing by 17, we see $x^2 + y^2 = \frac{16}{17}$. The center is $(0, 0)$ and the radius is $r = \frac{4}{\sqrt{17}}$.

The problem states that $L$ touches this circle. The perpendicular distance from the origin to $L$ must equal $r$. Using the distance formula $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$, we set up:

$$\frac{|-8|}{\sqrt{(3b + 4)^2 + (4 - 3b)^2}} = \frac{4}{\sqrt{17}}$$

Look at the denominator—it is a beautiful algebraic trap. Expanding $(3b + 4)^2$ and $(4 - 3b)^2$, we get $9b^2 + 24b + 16$ and $16 - 24b + 9b^2$. The $24b$ terms vanish into thin air!

We are left with $\sqrt{18b^2 + 32}$. Our equation becomes:

$$\frac{8}{\sqrt{18b^2 + 32}} = \frac{4}{\sqrt{17}}$$

Dividing by 4, we get $\frac{2}{\sqrt{18b^2 + 32}} = \frac{1}{\sqrt{17}}$. Squaring both sides yields:

$$\frac{4}{18b^2 + 32} = \frac{1}{17}$$

Cross-multiplying gives $68 = 18b^2 + 32$, so $18b^2 = 36$, which means **$b^2 = 2$**.

### The Final Ellipse

We have conquered the parameter $b$. The ellipse is $\frac{x^2}{5} + \frac{y^2}{b^2} = 1$, which becomes:

$$\frac{x^2}{5} + \frac{y^2}{2} = 1$$

Since $5 > 2$, this is a horizontal ellipse with $a^2 = 5$ and $b^2 = 2$. The eccentricity $e$ is given by:

$$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{2}{5}} = \sqrt{\frac{3}{5}}$$

You have navigated the intersection, the tangency, and the ellipse properties with precision. This is the essence of JEE Advanced—connecting disparate concepts into one coherent, beautiful solution.

Select Answer:

Visualized Solution

Visualizing the Geometry

Family of Lines Concept

Applying the Point $(1, 1)$

Solving for $λ$

Refining the Line Equation

The Tangency Condition

Applying the Distance Formula

Simplifying the Denominator

Solving for $b^{2}$

The Ellipse Equation

Calculating Eccentricity

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Power of the Family of Lines

The Tangency Condition

The Final Ellipse

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Select Answer:

Visualized Solution

Visualizing the Geometry

Family of Lines Concept

Applying the Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1,1)

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Refining the Line Equation

The Tangency Condition

Applying the Distance Formula

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Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

The Ellipse Equation

Calculating Eccentricity

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Power of the Family of Lines

The Tangency Condition

The Final Ellipse

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