The locus of the point of intersection of the lines $(\sqrt{3})kx + ky - 4\sqrt{3} = 0$ and $\sqrt{3}x - y - 4(\sqrt{3}) k = 0$ is a conic, whose eccentricity is

### Analyzing the Setup Imagine you are standing in a vast coordinate plane. You see two lines dancing across the grid governed by a parameter $k$. As $k$ changes, the lines shift, rotate, and slide, but they always intersect at a specific point. Our mission is to track the ghost of that intersection point. We seek the path it traces, which is the essence of finding a **locus** in coordinate geometry. ### The Bridge of Parameters We are given two equations: $$(\sqrt{3})kx + ky - 4\sqrt{3} = 0$$ $$\sqrt{3}x - y - 4(\sqrt{3})k = 0$$ Think of $k$ as a 'hidden variable' or the puppet master controlling the position of these lines. To find the locus, we must strip away the influence of $k$ by expressing it in terms of $x$ and $y$ from one equation and substituting it into the other. From the first equation, we factor out $k$: $$k(\sqrt{3}x + y) = 4\sqrt{3}$$ This yields our bridge equation: $$k = \frac{4\sqrt{3}}{\sqrt{3}x + y}$$ ### The Algebraic Symphony Now, we substitute this expression for $k$ into our second equation, $\sqrt{3}x - y = 4\sqrt{3}k$: $$\sqrt{3}x - y = 4\sqrt{3} \left( \frac{4\sqrt{3}}{\sqrt{3}x + y} \right)$$ Multiplying both sides by the denominator $(\sqrt{3}x + y)$, we obtain: $$(\sqrt{3}x - y)(\sqrt{3}x + y) = 4\sqrt{3} \times 4\sqrt{3}$$ Applying the difference of squares identity, $(a-b)(a+b) = a^2 - b^2$, the left side simplifies to $(\sqrt{3}x)^2 - y^2 = 3x^2 - y^2$. On the right side, we calculate $4\sqrt{3} \times 4\sqrt{3} = 16 \times 3 = 48$. Thus, the equation of our path is: $$3x^2 - y^2 = 48$$ ### Identifying the Conic To reveal the true nature of this path, we divide by $48$ to reach the standard form: $$\frac{3x^2}{48} - \frac{y^2}{48} = 1$$ $$\frac{x^2}{16} - \frac{y^2}{48} = 1$$ This is the classic equation of a **hyperbola**, where $a^2 = 16$ and $b^2 = 48$. The eccentricity $e$ of a hyperbola is defined by the formula $e = \sqrt{1 + \frac{b^2}{a^2}}$. Substituting our values: $$e = \sqrt{1 + \frac{48}{16}} = \sqrt{1 + 3} = \sqrt{4} = 2$$ ### Final Reflection We started with two shifting lines and ended with a fixed, elegant curve. The **eccentricity of 2** confirms the specific geometry of this hyperbola. In JEE Advanced mathematics, the key is to treat parameters like $k$ as variables to be eliminated. By doing so, you allow the underlying symmetry of the geometry to reveal itself.

JEE Main 2021 (February)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: The locus of the point of intersection of the lines $(3) k x + k y - 43 = 0$ and $3 x - y - 4 (3) k = 0$ is a conic, whose eccentricity is

Q: Wait, why do we eliminate $k$ instead of solving for $x$ and $y$ directly?

Great question! If you solve for $x$ and $y$ directly, you get specific values, but the problem asks for the *locus*—the entire path traced by the intersection point as $k$ varies. By eliminating $k$, we find the relationship that $x$ and $y$ must satisfy regardless of what $k$ is.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: The Spark. You are looking for the locus of a point

(x, y)

that satisfies two equations simultaneously. The variable

k

is a parameter that changes as the lines move. To find the path (locus) of the intersection, you must eliminate

k

entirely. Think of

k

as a bridge between the two equations that needs to be burned.

Hint 2

Hint 2: The Arsenal. Once you isolate

k

from the first equation, you have an expression in terms of

x

and

y

. When you substitute this into the second equation, look closely at the structure of the resulting expression. Does it resemble a classic algebraic identity like the difference of squares?

Hint 3

Hint 3: The Execution. Start by writing the first equation as

k (3 x + y) = 43

. Once you substitute

k

into the second equation, you will arrive at the form

(3 x - y) (3 x + y) = 43 \cdot 43

. Simplify this to reach the standard form of a conic section.

Visualized Solution

Analyze the Given Equations

Given Line 1: $(3) k x + k y - 43 = 0$
Given Line 2: $3 x - y - 43 k = 0$
Goal: Find the locus of the intersection point by eliminating the parameter $k$ .

Identify the Parameter $k$

Identify the parameter to eliminate: $k$
Look at Line 1 to isolate $k$ .

Isolate $k$ from Line 1

From Line 1: $k (3 x + y) = 43$
Isolating $k$ : $k = \frac{4 3}{3 x + y}$

Substitute $k$ into Line 2

Substitute $k$ into Line 2: $3 x - y = 43 k$
$3 x - y = 43 (\frac{4 3}{3 x + y})$

Cross-Multiply the Terms

Multiply both sides by $(3 x + y)$ :
$(3 x - y) (3 x + y) = 43 \times 43$

Apply Difference of Squares

Using $(a - b) (a + b) = a^{2} - b^{2}$ :
$(3 x)^{2} - (y)^{2} = 16 \times 3$
$3 x^{2} - y^{2} = 48$

Convert to Standard Form

Divide the equation by $48$ :
$\frac{3 x ^{2}}{48} - \frac{y ^{2}}{48} = 1$
$\frac{x ^{2}}{16} - \frac{y ^{2}}{48} = 1$

Identify the Conic Type

The equation is of the form $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$
This represents a Hyperbola.
Comparing terms: $a^{2} = 16$ and $b^{2} = 48$

Eccentricity Formula

For a hyperbola, eccentricity $e = 1 + \frac{b ^{2}}{a ^{2}}$
Substitute $a^{2} = 16$ and $b^{2} = 48$ into the formula.
$e = 1 + \frac{48}{16}$

Final Calculation

$e = 1 + 3$
$e = 4$
$e = 2$
The eccentricity of the conic is $2$ .

00:00 / 00:00

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

Analyzing the Setup

Imagine you are standing in a vast coordinate plane. You see two lines dancing across the grid governed by a parameter

k

. As

k

changes, the lines shift, rotate, and slide, but they always intersect at a specific point.

Our mission is to track the ghost of that intersection point. We seek the path it traces, which is the essence of finding a locus in coordinate geometry.

The Bridge of Parameters

We are given two equations:

(3) k x + k y - 43 = 0

3 x - y - 4 (3) k = 0

Think of

k

as a 'hidden variable' or the puppet master controlling the position of these lines. To find the locus, we must strip away the influence of

k

by expressing it in terms of

x

and

y

from one equation and substituting it into the other.

From the first equation, we factor out

k

k (3 x + y) = 43

This yields our bridge equation:

k = \frac{4 3}{3 x + y}

The Algebraic Symphony

Now, we substitute this expression for

k

into our second equation,

3 x - y = 43 k

3 x - y = 43 (\frac{4 3}{3 x + y})

Multiplying both sides by the denominator

(3 x + y)

, we obtain:

(3 x - y) (3 x + y) = 43 \times 43

Applying the difference of squares identity,

(a - b) (a + b) = a^{2} - b^{2}

, the left side simplifies to

(3 x)^{2} - y^{2} = 3 x^{2} - y^{2}

. On the right side, we calculate

43 \times 43 = 16 \times 3 = 48

Thus, the equation of our path is:

3 x^{2} - y^{2} = 48

Identifying the Conic

To reveal the true nature of this path, we divide by

48

to reach the standard form:

\frac{3 x ^{2}}{48} - \frac{y ^{2}}{48} = 1

\frac{x ^{2}}{16} - \frac{y ^{2}}{48} = 1

This is the classic equation of a hyperbola, where

a^{2} = 16

and

b^{2} = 48

. The eccentricity

e

of a hyperbola is defined by the formula

e = 1 + \frac{b ^{2}}{a ^{2}}

Substituting our values:

e = 1 + \frac{48}{16} = 1 + 3 = 4 = 2

Final Reflection

We started with two shifting lines and ended with a fixed, elegant curve. The eccentricity of 2 confirms the specific geometry of this hyperbola.

In JEE Advanced mathematics, the key is to treat parameters like

k

as variables to be eliminated. By doing so, you allow the underlying symmetry of the geometry to reveal itself.

Common Traps & FAQs

Wait, why do we eliminate

k

instead of solving for

x

and

y

directly?

Great question! If you solve for

x

and

y

directly, you get specific values, but the problem asks for the *locus*—the entire path traced by the intersection point as

k

varies. By eliminating

k

, we find the relationship that

x

and

y

must satisfy regardless of what

k

is.

How do I know if the resulting equation is a hyperbola or an ellipse?

Look at the signs in your final equation. If you have a minus sign between the

x^{2}

and

y^{2}

terms (like

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

), it is a hyperbola. If it were a plus sign, it would be an ellipse. The structure of the difference of squares here naturally leads to a hyperbola.

Is there a shortcut to finding the eccentricity without converting to standard form?

While you could memorize properties of specific conic forms, converting to the standard form

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

is the safest and most robust way to avoid errors. It ensures you correctly identify

a^{2}

and

b^{2}

before plugging them into

e = 1 + \frac{b ^{2}}{a ^{2}}

Why is the eccentricity of a hyperbola always greater than 1?

Eccentricity measures how 'stretched' a conic is. For a hyperbola, the curve opens outward, meaning it is 'more than' a circle or ellipse. Mathematically, since

e = 1 + \frac{b ^{2}}{a ^{2}}

, and both

a^{2}

and

b^{2}

are positive, the value inside the square root must be greater than 1.

What if I made a mistake in the cross-multiplication step?

Algebraic errors are the most common traps in JEE! Always double-check your distribution. In this problem, the product

(3 x - y) (3 x + y)

is a classic

(a - b) (a + b)

pattern. If you expand it carefully, you'll see the middle terms cancel out perfectly, leaving you with

3 x^{2} - y^{2}

Animated Solution for Mathematics - Conic Sections: The locus of the point of intersection of the lines $(3) k x + k y - 43 = 0$ and $3 x - y - 4 (3) k = 0$ is a conic, whose eccentricity is

Enter Numerical Value:

Visualized Solution

Analyze the Given Equations

Identify the Parameter $k$

Isolate $k$ from Line 1

Substitute $k$ into Line 2

Cross-Multiply the Terms

Apply Difference of Squares

Convert to Standard Form

Identify the Conic Type

Eccentricity Formula

Final Calculation

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

Analyzing the Setup

The Bridge of Parameters

The Algebraic Symphony

Identifying the Conic

Final Reflection

Similar Questions

The locus of the point of intersection of the lines, $2 x - y + 42 k = 0$ and $2 k x + k y - 42 = 0$ (k is any non-zero real parameter), is :

The locus of the mid point of the line segment joining the point $(4, 3)$ and the points on the ellipse $x^{2} + 2 y^{2} = 4$ is an ellipse with eccentricity :

Match the conics in Column I with the statements/expressions in Column II.

List-I

List-II

A hyperbola has its centre at the origin, passes through the point (4,2) and has transverse axis of length 4 along the x-axis. Then the eccentricity of the hyperbola is :

The locus of a variable point whose distance from $(2, 0)$ is $2/3$ times its distance from the line $x = - 9/2$ is

The ellipse $E_{1} : \frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes. Another ellipse $E_{2}$ passing through the point $(0, 4)$ circumscribes the rectangle $R$ . The eccentricity of the ellipse $E_{2}$ is

Enter Numerical Value:

Visualized Solution

Analyze the Given Equations

Identify the Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Isolate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k from Line 1

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k into Line 2

Cross-Multiply the Terms

Apply Difference of Squares

Convert to Standard Form

Identify the Conic Type

Eccentricity Formula

Final Calculation

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

Analyzing the Setup

The Bridge of Parameters

The Algebraic Symphony

Identifying the Conic

Final Reflection

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the point of intersection of the lines, 2​x−y+42​k=0 and 2​kx+ky−42​=0 (k is any non-zero real parameter), is :

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Match the conics in Column I with the statements/expressions in Column II.

List-I

List-II

A hyperbola has its centre at the origin, passes through the point (4,2) and has transverse axis of length 4 along the x-axis. Then the eccentricity of the hyperbola is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of a variable point whose distance from (2,0) is 2/3 times its distance from the line x=−9/2 is

Identify the Parameter $k$

Isolate $k$ from Line 1

Substitute $k$ into Line 2

The locus of the point of intersection of the lines, $2 x - y + 42 k = 0$ and $2 k x + k y - 42 = 0$ (k is any non-zero real parameter), is :

The locus of the mid point of the line segment joining the point $(4, 3)$ and the points on the ellipse $x^{2} + 2 y^{2} = 4$ is an ellipse with eccentricity :

The locus of a variable point whose distance from $(2, 0)$ is $2/3$ times its distance from the line $x = - 9/2$ is