Why can we set the chord equation to $lx + my = 1$?

Because the chord does not pass through the origin (if it did, it wouldn't be a chord subtending a right angle at the origin in the standard sense, or it would be a degenerate case). This form allows us to homogenize the equation, which is a powerful technique for dealing with lines passing through the origin.

Why does $A+B=0$ work for perpendicular lines?

The angle $\theta$ between lines $Ax^2 + 2Hxy + By^2 = 0$ is given by $\tan \theta = \frac{2\sqrt{H^2 - AB}}{A+B}$. For $\theta = 90^\circ$, $\tan \theta$ is undefined, implying the denominator $A+B$ must be zero.

Is the fixed point unique?

Yes, the family of lines $(x-1) + m(2x+y) = 0$ represents all lines passing through the intersection of $x-1=0$ and $2x+y=0$, which is a single unique point.

JEE Advanced 1991

LEVELJEE Advanced

Animated Solution for Mathematics - Straight Lines: Show that all chords of the curve $3 x^{2} - y^{2} - 2 x + 4 y = 0$ , which subtend a right angle at the origin, pass through a fixed point. Find the coordinates of the point.

Visualized Solution

Visualizing the Curve and Chord

Given second-degree curve: $3 x^{2} - y^{2} - 2 x + 4 y = 0$
The curve passes through the origin $(0, 0)$ because substituting $x = 0, y = 0$ satisfies the equation.
We need to analyze chords $A B$ that subtend a right angle ( $9 0^{\circ}$ ) at the origin.

Equation of the Dynamic Chord

Let the equation of the chord $A B$ be $l x + m y = 1$ .
Here, $l$ and $m$ are variable parameters that change as the chord moves.
Writing it as $l x + m y = 1$ allows us to homogenize the curve's equation.

Homogenizing the Curve Equation

To find the joint equation of lines $O A$ and $O B$ , we homogenize the curve equation.
We multiply the linear terms $- 2 x + 4 y$ by $(l x + m y)$ which is equal to $1$ .
The homogenized equation is: $3 x^{2} - y^{2} - (2 x - 4 y) (l x + m y) = 0$ .

Expanding and Grouping Terms

Let's expand the expression:
$3 x^{2} - y^{2} - (2 l x^{2} + 2 m x y - 4 l x y - 4 m y^{2}) = 0$
Grouping the $x^{2}$ , $y^{2}$ , and $x y$ terms together:
$(3 - 2 l) x^{2} + (4 l - 2 m) x y + (4 m - 1) y^{2} = 0$

Condition for Perpendicular Lines

For the pair of lines $A x^{2} + 2 H x y + B y^{2} = 0$ to be perpendicular:
$Coefficient of x^{2} + Coefficient of y^{2} = 0$
Therefore, $A + B = 0$ .

Finding the Relation between $l$ and $m$

Substitute the coefficients into the perpendicularity condition:
$(3 - 2 l) + (4 m - 1) = 0$
Simplify the equation:
$2 - 2 l + 4 m = 0 ⟹ l - 2 m = 1 ⟹ l = 1 + 2 m$

Rewriting the Chord Equation

Substitute $l = 1 + 2 m$ back into the chord equation $l x + m y = 1$ :
$(1 + 2 m) x + m y = 1$
Expand and rearrange terms:
$(x - 1) + m (2 x + y) = 0$

Finding the Fixed Point $(1, - 2)$

The equation $(x - 1) + m (2 x + y) = 0$ represents a family of lines passing through the intersection of:
$x - 1 = 0 ⟹ x = 1$
$2 x + y = 0 ⟹ y = - 2 x$
Substituting $x = 1$ into $2 x + y = 0$ gives $y = - 2$ .
Therefore, all such chords pass through the fixed point $(1, - 2)$ .

00:00 / 00:00

The Sigma Insight: Family of Lines

Solution Diagram

Analyzing the Setup

The given curve is defined by the equation:

3 x^{2} - y^{2} - 2 x + 4 y = 0

We consider a chord

A B

represented by the linear equation:

l x + m y = 1

Our goal is to prove that if the lines

O A

and

O B

(connecting the origin to the chord endpoints) are perpendicular, the chord

A B

must pass through a fixed point.

The Magic of Homogenization

To analyze the lines

O A

and

O B

, we homogenize the equation of the curve using the chord equation. We rewrite the linear terms of the curve equation by multiplying them by

(l x + m y)

, which is equal to

1

3 x^{2} - y^{2} - (2 x - 4 y) (l x + m y) = 0

Expanding this expression, we obtain:

3 x^{2} - y^{2} - (2 l x^{2} + 2 m x y - 4 l x y - 4 m y^{2}) = 0

Grouping the terms by their variables, we arrive at the joint equation for the pair of lines:

(3 - 2 l) x^{2} + (4 l - 2 m) x y + (4 m - 1) y^{2} = 0

The Perpendicularity Condition

For a homogeneous equation of the form

A x^{2} + 2 H x y + B y^{2} = 0

, the lines are perpendicular if and only if the sum of the coefficients of

x^{2}

and

y^{2}

is zero, i.e.,

A + B = 0

Applying this condition to our joint equation:

(3 - 2 l) + (4 m - 1) = 0

Simplifying this expression, we get:

2 - 2 l + 4 m = 0 \Rightarrow l - 2 m = 1

The Final Reveal

We now substitute

l = 1 + 2 m

back into the original chord equation

l x + m y = 1

(1 + 2 m) x + m y = 1

Rearranging the terms to isolate the parameter

m

, we get:

(x - 1) + m (2 x + y) = 0

This equation represents a family of lines passing through the intersection of the lines

x - 1 = 0

and

2 x + y = 0

. Solving these simultaneously:

1. From

x - 1 = 0

, we find

x = 1

. 2. Substituting

x = 1

into

2 x + y = 0

, we find

2 (1) + y = 0

, which gives

y = - 2

Thus, all such chords must pass through the fixed point $(1, - 2)$ .

Animated Solution for Mathematics - Straight Lines: Show that all chords of the curve $3 x^{2} - y^{2} - 2 x + 4 y = 0$ , which subtend a right angle at the origin, pass through a fixed point. Find the coordinates of the point.

Visualized Solution

Visualizing the Curve and Chord

Equation of the Dynamic Chord

Homogenizing the Curve Equation

Expanding and Grouping Terms

Condition for Perpendicular Lines

Finding the Relation between $l$ and $m$

Rewriting the Chord Equation

Finding the Fixed Point $(1, - 2)$

The Sigma Insight: Family of Lines

Analyzing the Setup

The Magic of Homogenization

The Perpendicularity Condition

The Final Reveal

Similar Questions

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The set of lines $a x + b y + c = 0$ , where $3 a + 2 b + 4 c = 0$ is concurrent at the point .........

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Lines $L_{1} \equiv a x + b y + c = 0$ and $L_{2} \equiv l x + m y + n = 0$ intersect at the point $P$ and make an angle $θ$ with each other. Find the equation of a line $L$ different from $L_{2}$ which passes through $P$ and makes the same angle $θ$ with $L_{1}$ .

Three lines $p x + q y + r = 0, q x + r y + p = 0$ and $r x + p y + q = 0$ are concurrent if

Visualized Solution

Visualizing the Curve and Chord

Equation of the Dynamic Chord

Homogenizing the Curve Equation

Expanding and Grouping Terms

Condition for Perpendicular Lines

Finding the Relation between .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Rewriting the Chord Equation

Finding the Fixed Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1,−2)

The Sigma Insight: Family of Lines

Analyzing the Setup

The Magic of Homogenization

The Perpendicularity Condition

The Final Reveal

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The set of lines $a x + b y + c = 0$ , where $3 a + 2 b + 4 c = 0$ is concurrent at the point .........

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