JEE Main 2025 (January)

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: Let the circle C touch the line $x - y + 1 = 0$ , have the centre on the positive x -axis, and cut off a chord of length $\frac{4}{13}$ along the line $- 3 x + 2 y = 1$ . Let H be the hyperbola $\frac{x ^{2}}{α ^{2}} - \frac{y ^{2}}{β ^{2}} = 1$ whose one of the foci is the centre of C and the length of the transverse axis is the diameter of C. Then $2 α^{2} + 3 β^{2}$ is equal to

Q: Why do we reject the negative root when solving the quadratic equation for $a$?

Excellent catch! When we solve the quadratic $5a^2 - 14a - 3 = 0$, we get two roots: $a = 3$ and $a = -1/5$. However, the problem explicitly states that the center of the circle lies on the *positive* x-axis. This constraint forces us to discard the negative value, as it would place the center on the negative side of the axis.

Enter Numerical Value:

Progressive Hints

Hint 1

Visualize the circle's center at

(a, 0)

on the x-axis. The condition that the circle touches the line

x - y + 1 = 0

implies that the perpendicular distance from the center

(a, 0)

to this line is exactly equal to the radius

r

. Use the point-to-line distance formula to express

r

in terms of

a

Hint 2

The circle cuts a chord of length

\frac{4}{13}

on the line

- 3 x + 2 y = 1

. Recall that the perpendicular distance

d

from the center to a chord, the radius

r

, and half the chord length

L /2

form a right-angled triangle. Use the Pythagorean theorem:

r^{2} = d^{2} + (L /2)^{2}

Hint 3

Substitute your expression for

r

(from Hint 1) and the distance

d

(calculated from the center to the chord line) into the equation

r^{2} - d^{2} = (L /2)^{2}

. This will yield a quadratic equation in

a

. Solve for

a

, keeping in mind the constraint that the center lies on the positive x-axis.

Visualized Solution

Defining the Center of Circle $C$

Let the center of circle $C$ be $(a, 0)$ where $a > 0$ .
The circle lies on the positive x-axis as per the problem constraints.

Radius from Tangency Condition

Circle $C$ touches the line $x - y + 1 = 0$ .
Radius $r = Perpendicular distance from (a, 0) to x - y + 1 = 0$
$r = \frac{∣ a - 0 + 1∣}{1 ^{2} + ( - 1 ) ^{2}} = \frac{a + 1}{2}$

Distance to the Chord Line

Distance $d$ from center $(a, 0)$ to the line $- 3 x + 2 y - 1 = 0$ :
$d = \frac{∣ - 3 a + 2 ( 0 ) - 1∣}{( - 3 ) ^{2} + 2 ^{2}} = \frac{3 a + 1}{13}$

Using the Chord Length Formula

Length of chord $= 2 r^{2} - d^{2} = \frac{4}{13}$
Squaring both sides: $r^{2} - d^{2} = (\frac{2}{13})^{2} = \frac{4}{13}$

Substituting $r$ and $d$

Substitute $r = \frac{a + 1}{2}$ and $d = \frac{3 a + 1}{13}$ into the equation:
$(\frac{a + 1}{2})^{2} - (\frac{3 a + 1}{13})^{2} = \frac{4}{13}$

Expanding the Terms

Expand the squares:
$\frac{a ^{2} + 2 a + 1}{2} - \frac{9 a ^{2} + 6 a + 1}{13} = \frac{4}{13}$

Simplifying to a Quadratic

Multiply by 26 to clear denominators:
$13 (a^{2} + 2 a + 1) - 2 (9 a^{2} + 6 a + 1) = 8$
$13 a^{2} + 26 a + 13 - 18 a^{2} - 12 a - 2 = 8$
$- 5 a^{2} + 14 a + 11 = 8 \Rightarrow 5 a^{2} - 14 a - 3 = 0$

Solving for $a$

Factorize the quadratic equation:
$(5 a + 1) (a - 3) = 0$
Since $a > 0$ , we have $a = 3$ .
Center of Circle $C$ : $(3, 0)$

Finding Circle Radius and Diameter

Radius $r = \frac{3 + 1}{2} = \frac{4}{2} = 22$
Diameter of circle $C = 2 r = 42$

Hyperbola Parameters: $α$ and Focus

For Hyperbola $H : \frac{x ^{2}}{α ^{2}} - \frac{y ^{2}}{β ^{2}} = 1$
Transverse axis $2 α = Diameter = 42 \Rightarrow α = 22 \Rightarrow α^{2} = 8$
One focus is $(3, 0) \Rightarrow α e = 3$

Calculating $β^{2}$

Using the relation $β^{2} = α^{2} (e^{2} - 1) = (α e)^{2} - α^{2}$ :
$β^{2} = 3^{2} - 8 = 9 - 8 = 1$

Final Calculation

Calculate the final expression:
$2 α^{2} + 3 β^{2} = 2 (8) + 3 (1)$
$= 16 + 3 = 19$

00:00 / 00:00

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

Solution Diagram

Analyzing the Setup

We begin with a circle

C

whose center lies on the positive

x

-axis. Let us denote this center as

(a, 0)

, where

a > 0

The circle touches the line

x - y + 1 = 0

. In geometry, when a circle touches a line, that line is a tangent.

The perpendicular distance from the center

(a, 0)

to the line

x - y + 1 = 0

must be equal to the radius

r

. Using the distance formula, we find:

r = \frac{∣ a - 0 + 1∣}{1 ^{2} + ( - 1 ) ^{2}} = \frac{a + 1}{2}

The Chord's Secret

The circle cuts a chord of length

\frac{4}{13}

on the line

- 3 x + 2 y = 1

. If we draw a perpendicular from the center to this chord, it bisects the chord, creating a right-angled triangle where the radius is the hypotenuse.

The distance

d

from the center

(a, 0)

to the line

- 3 x + 2 y - 1 = 0

is:

d = \frac{∣ - 3 a + 2 ( 0 ) - 1∣}{( - 3 ) ^{2} + 2 ^{2}} = \frac{3 a + 1}{13}

The chord length formula is

L = 2 r^{2} - d^{2}

. Given

L = \frac{4}{13}

, we have

L /2 = \frac{2}{13}

. Squaring both sides, we obtain:

r^{2} - d^{2} = \frac{4}{13}

The Algebraic Bridge

We substitute our expressions for

r

and

d

into the equation:

(\frac{a + 1}{2})^{2} - (\frac{3 a + 1}{13})^{2} = \frac{4}{13}

Expanding this, we get:

\frac{a ^{2} + 2 a + 1}{2} - \frac{9 a ^{2} + 6 a + 1}{13} = \frac{4}{13}

Multiplying by

26

to clear the denominators, we arrive at:

13 (a^{2} + 2 a + 1) - 2 (9 a^{2} + 6 a + 1) = 8

This simplifies to

13 a^{2} + 26 a + 13 - 18 a^{2} - 12 a - 2 = 8

, which reduces to the quadratic:

5 a^{2} - 14 a - 3 = 0

Factoring this, we get

(5 a + 1) (a - 3) = 0

. Since

a > 0

, we must have $a = 3$ . The center is

(3, 0)

The Hyperbola's Entrance

With

a = 3

, the radius

r = \frac{3 + 1}{2} = 22

. The diameter is

42

The hyperbola

H

has a transverse axis of length

2 α

, which equals the diameter. So,

2 α = 42

, meaning

α = 22

and

α^{2} = 8

The focus is the center of the circle,

(3, 0)

, so

α e = 3

. We need

β^{2}

. Using the hyperbola identity

β^{2} = α^{2} (e^{2} - 1) = (α e)^{2} - α^{2}

, we calculate:

β^{2} = 3^{2} - 8 = 9 - 8 = 1

Finally, the expression

2 α^{2} + 3 β^{2}

is:

2 (8) + 3 (1) = 16 + 3 = 19

The final answer is $19$ .

Common Traps & FAQs

Wait, why do we use the perpendicular distance formula for the radius? Isn't the radius just the distance to the origin?

That is a common confusion! The radius is the distance from the center to *any* point on the circumference. When a line is tangent to a circle, the radius drawn to the point of tangency is perpendicular to that line. Therefore, the shortest distance from the center to the tangent line is exactly the radius. It is a geometric property, not a coordinate-based one!

Why do we reject the negative root when solving the quadratic equation for

a

Excellent catch! When we solve the quadratic

5 a^{2} - 14 a - 3 = 0

, we get two roots:

a = 3

and

a = - 1/5

. However, the problem explicitly states that the center of the circle lies on the *positive* x-axis. This constraint forces us to discard the negative value, as it would place the center on the negative side of the axis.

How do I relate the hyperbola's focus to the circle's center?

The problem states that one of the foci of the hyperbola is the center of the circle. Since the center of the circle is

(3, 0)

, the focus of the hyperbola is also at

(3, 0)

. For a standard hyperbola

\frac{x ^{2}}{α ^{2}} - \frac{y ^{2}}{β ^{2}} = 1

, the focus is at

(α e, 0)

. Thus,

α e = 3

. This is the bridge between the two curves!

Enter Numerical Value:

Visualized Solution

Defining the Center of Circle $C$

Radius from Tangency Condition

Distance to the Chord Line

Using the Chord Length Formula

Substituting $r$ and $d$

Expanding the Terms

Simplifying to a Quadratic

Solving for $a$

Finding Circle Radius and Diameter

Hyperbola Parameters: $α$ and Focus

Calculating $β^{2}$

Final Calculation

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

Analyzing the Setup

The Chord's Secret

The Algebraic Bridge

The Hyperbola's Entrance

Similar Questions

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Comprehension Passage

Equation of a common tangent with positive slope to the circle as well as to the hyperbola is

Equation of the circle with $A B$ as its diameter is

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Enter Numerical Value:

Visualized Solution

Defining the Center of Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Radius from Tangency Condition

Distance to the Chord Line

Using the Chord Length Formula

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Expanding the Terms

Simplifying to a Quadratic

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

Finding Circle Radius and Diameter

Hyperbola Parameters: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α and Focus

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β2

Final Calculation

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

Analyzing the Setup

The Chord's Secret

The Algebraic Bridge

The Hyperbola's Entrance

Similar Questions

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