The locus of the point of intersection of the lines, $\sqrt{2}x - y + 4\sqrt{2}k = 0$ and $\sqrt{2}kx + ky - 4\sqrt{2} = 0$ (k is any non-zero real parameter), is : Options: (A) a hyperbola with length of its transverse axis $8\sqrt{2}$ (B) a hyperbola whose eccentricity is $\sqrt{3}$ (C) an ellipse whose eccentricity is $\frac{1}{\sqrt{3}}$ (D) an ellipse with length of its major axis $8\sqrt{2}$

### Analyzing the Setup We are given two lines that depend on a parameter $k$: 1) $\sqrt{2}x - y + 4\sqrt{2}k = 0$ 2) $\sqrt{2}kx + ky - 4\sqrt{2} = 0$ Our objective is to find the locus of the point of intersection $(x, y)$ as $k$ varies. ### The Strategic Pivot Instead of solving for $k$ and substituting, we isolate the parameter to eliminate it. Rewrite the equations as follows: 1) $\sqrt{2}x - y = -4\sqrt{2}k$ 2) $k(\sqrt{2}x + y) = 4\sqrt{2}$ By multiplying these two equations, the parameter $k$ will cancel out, leaving us with a direct relationship between $x$ and $y$. ### The Algebraic Symphony Multiplying the two equations yields: $$(\sqrt{2}x - y) \cdot k(\sqrt{2}x + y) = (-4\sqrt{2}k) \cdot (4\sqrt{2})$$ Assuming $k \neq 0$, we divide both sides by $k$: $$(\sqrt{2}x - y)(\sqrt{2}x + y) = -16 \cdot 2$$ $$(\sqrt{2}x - y)(\sqrt{2}x + y) = -32$$ Applying the difference of squares identity $(a-b)(a+b) = a^2 - b^2$, we obtain: $$(\sqrt{2}x)^2 - y^2 = -32$$ $$2x^2 - y^2 = -32$$ ### Identifying the Geometric Soul To identify the curve, we rearrange the equation into standard form by dividing by $-32$: $$\frac{2x^2}{-32} - \frac{y^2}{-32} = 1$$ $$\frac{y^2}{32} - \frac{x^2}{16} = 1$$ This is the equation of a **hyperbola** opening along the $y$-axis. Comparing this to the standard form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, we identify $a^2 = 32$, which implies $a = 4\sqrt{2}$. The length of the transverse axis is given by $2a$: $$2a = 2(4\sqrt{2}) = 8\sqrt{2}$$ The locus is a **hyperbola** with a transverse axis of length **$8\sqrt{2}$**.

JEE Main 2018 (16 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: 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 :

Q: How do I know it's a hyperbola and not an ellipse?

Great question! Look at the final equation: $\frac{y^2}{32} - \frac{x^2}{16} = 1$. The key is the sign between the squared terms. If it were an ellipse, the signs would be the same (both positive). Because one is positive and one is negative, it represents a hyperbola. The negative sign is the signature of a hyperbola!

Q: What happens if $k=0$?

The problem explicitly states that $k$ is a non-zero real parameter. This is a crucial piece of information! If $k$ were zero, the equations would collapse into something else entirely, and the intersection point might not even exist or would be undefined. By defining $k \neq 0$, the problem ensures that the lines are always well-defined and intersect, allowing us to trace the locus.

Select Answer:

Progressive Hints

Hint 1

Imagine the two lines as two dancers on a stage. As the parameter

k

changes, the lines rotate and shift. Your task is to find the path traced by their intersection point. Don't get lost in the individual positions; focus on the relationship that remains invariant regardless of

k

Hint 2

Resist the urge to solve for

x

and

y

in terms of

k

and substitute. That path is a labyrinth of algebra. Instead, look at the structure of the two equations. Can you isolate the terms involving

k

and the terms that are constant? Think about how multiplication might help you eliminate

k

entirely.

Hint 3

Start by rewriting the equations as

2 x - y = - 42 k

and

k (2 x + y) = 42

. Now, multiply the left-hand sides and the right-hand sides together. Watch how the

k

terms interact and cancel out, leaving you with a beautiful, parameter-free equation.

Visualized Solution

Analyze the Given Equations

Given Line 1: $2 x - y + 42 k = 0$
Given Line 2: $2 k x + k y - 42 = 0$
Goal: Eliminate the parameter $k$ to find the locus of their intersection.

Rearrange Equations to Isolate $k$

Rearranging Line 1: $2 x - y = - 42 k$ ... (1)
Rearranging Line 2: $k (2 x + y) = 42$ ... (2)

Eliminate the Parameter $k$

Multiply Equation (1) and (2):
$k (2 x - y) (2 x + y) = (- 42 k) (42)$

Cancel $k$ and Simplify Constants

Since $k \neq = 0$ , divide both sides by $k$ :
$(2 x - y) (2 x + y) = - 16 \cdot 2$
$(2 x - y) (2 x + y) = - 32$

Apply Algebraic Identity

Using $(a - b) (a + b) = a^{2} - b^{2}$ :
$(2 x)^{2} - y^{2} = - 32$
$2 x^{2} - y^{2} = - 32$

Convert to Standard Form

Divide the entire equation by $- 32$ :
$\frac{2 x ^{2}}{- 32} - \frac{y ^{2}}{- 32} = 1$
$\frac{y ^{2}}{32} - \frac{x ^{2}}{16} = 1$

Identify the Locus

Standard form: $\frac{y ^{2}}{32} - \frac{x ^{2}}{16} = 1$
This represents a Hyperbola opening along the y-axis.
Comparing with $\frac{y ^{2}}{a ^{2}} - \frac{x ^{2}}{b ^{2}} = 1$ :
$a^{2} = 32 ⟹ a = 42$
$b^{2} = 16 ⟹ b = 4$

Calculate Transverse Axis Length

Length of Transverse Axis $= 2 a$
$= 2 (42)$
$= 82$
This matches Option (a).

Verify Eccentricity

Eccentricity $e = 1 + \frac{b ^{2}}{a ^{2}}$
$e = 1 + \frac{16}{32} = 1 + \frac{1}{2}$
$e = \frac{3}{2}$
Since $e \neq = 3$ , Option (b) is incorrect.

Conclusion

Key Takeaway: Eliminate the parameter $k$ using algebraic manipulation to find the locus.
Final Result: The locus is a hyperbola with transverse axis length $82$ .

00:00 / 00:00

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

Solution Diagram

Analyzing the Setup

We are given two lines that depend on a parameter

k

: 1)

2 x - y + 42 k = 0

2 k x + k y - 42 = 0

Our objective is to find the locus of the point of intersection

(x, y)

k

varies.

The Strategic Pivot

Instead of solving for

k

and substituting, we isolate the parameter to eliminate it. Rewrite the equations as follows: 1)

2 x - y = - 42 k

k (2 x + y) = 42

By multiplying these two equations, the parameter

k

will cancel out, leaving us with a direct relationship between

x

and

y

The Algebraic Symphony

Multiplying the two equations yields:

(2 x - y) \cdot k (2 x + y) = (- 42 k) \cdot (42)

Assuming $k eq 0$, we divide both sides by

k

(2 x - y) (2 x + y) = - 16 \cdot 2

(2 x - y) (2 x + y) = - 32

Applying the difference of squares identity

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

, we obtain:

(2 x)^{2} - y^{2} = - 32

2 x^{2} - y^{2} = - 32

Identifying the Geometric Soul

To identify the curve, we rearrange the equation into standard form by dividing by

- 32

\frac{2 x ^{2}}{- 32} - \frac{y ^{2}}{- 32} = 1

\frac{y ^{2}}{32} - \frac{x ^{2}}{16} = 1

This is the equation of a hyperbola opening along the

y

-axis. Comparing this to the standard form

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

, we identify

a^{2} = 32

, which implies

a = 42

The length of the transverse axis is given by

2 a

2 a = 2 (42) = 82

The locus is a hyperbola with a transverse axis of length $82$ .

Common Traps & FAQs

Why can't I just solve for

x

and

y

in terms of

k

and then substitute? It seems more direct.

That is a classic trap! While substitution is a valid algebraic method, it often leads to a nightmare of fractions and complex expressions that are prone to calculation errors. In JEE Advanced, we look for the 'elegant' path. By multiplying the equations, we use the structure of the problem to our advantage, bypassing the need to explicitly solve for

k

How do I know it's a hyperbola and not an ellipse?

Great question! Look at the final equation:

\frac{y ^{2}}{32} - \frac{x ^{2}}{16} = 1

. The key is the sign between the squared terms. If it were an ellipse, the signs would be the same (both positive). Because one is positive and one is negative, it represents a hyperbola. The negative sign is the signature of a hyperbola!

What happens if

k = 0

The problem explicitly states that

k

is a non-zero real parameter. This is a crucial piece of information! If

k

were zero, the equations would collapse into something else entirely, and the intersection point might not even exist or would be undefined. By defining

k \neq = 0

, the problem ensures that the lines are always well-defined and intersect, allowing us to trace the locus.

Animated Solution for Mathematics - Conic Sections: 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 :

Select Answer:

Visualized Solution

Analyze the Given Equations

Rearrange Equations to Isolate $k$

Eliminate the Parameter $k$

Cancel $k$ and Simplify Constants

Apply Algebraic Identity

Convert to Standard Form

Identify the Locus

Calculate Transverse Axis Length

Verify Eccentricity

Conclusion

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

Analyzing the Setup

The Strategic Pivot

The Algebraic Symphony

Identifying the Geometric Soul

Similar Questions

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

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 circle touches the x- axis and also touches the circle with centre at $(0, 3)$ and radius 2. The locus of the centre of the circle is

The locus of a point which divides the line segment joining the point $(0, - 1)$ and a point on the parabola, $x^{2} = 4 y$ , internally in the ratio 1: 2, 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

If $a \neq = 0$ and the line $2 b x + 3 cy + 4 d = 0$ passes through the points of intersection of the parabolas $y^{2} = 4 a x$ and $x^{2} = 4 a y$ , then

The equation $2 x^{2} + 3 y^{2} - 8 x - 18 y + 35 = k$ represents

A hyperbola, having the transverse axis of length $2 sin θ$ , is confocal with the ellipse $3 x^{2} + 4 y^{2} = 12$ . Then its equation is

A hyperbola having the transverse axis of length $2$ has the same foci as that of the ellipse of $3 x^{2} + 4 y^{2} = 12$ , then this hyperbola does not pass through which of the following points?

Select Answer:

Visualized Solution

Analyze the Given Equations

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

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

Cancel .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k and Simplify Constants

Apply Algebraic Identity

Convert to Standard Form

Identify the Locus

Calculate Transverse Axis Length

Verify Eccentricity

Conclusion

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

Analyzing the Setup

The Strategic Pivot

The Algebraic Symphony

Identifying the Geometric Soul

Similar Questions

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List-II

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Rearrange Equations to Isolate $k$

Eliminate the Parameter $k$

Cancel $k$ and Simplify Constants

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

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 :

A circle touches the x- axis and also touches the circle with centre at $(0, 3)$ and radius 2. The locus of the centre of the circle is

The locus of a point which divides the line segment joining the point $(0, - 1)$ and a point on the parabola, $x^{2} = 4 y$ , internally in the ratio 1: 2, 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

If $a \neq = 0$ and the line $2 b x + 3 cy + 4 d = 0$ passes through the points of intersection of the parabolas $y^{2} = 4 a x$ and $x^{2} = 4 a y$ , then

The equation $2 x^{2} + 3 y^{2} - 8 x - 18 y + 35 = k$ represents

A hyperbola, having the transverse axis of length $2 sin θ$ , is confocal with the ellipse $3 x^{2} + 4 y^{2} = 12$ . Then its equation is