JEE Advanced 2010

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: The line $2 x + y = 1$ is tangent to the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ . If this line passes through the point of intersection of the nearest directrix and the x-axis, then the eccentricity of the hyperbola is

Q: Can I solve this using the discriminant of the quadratic equation?

Yes, you absolutely can! If you substitute $y = 1 - 2x$ into the hyperbola equation and set the discriminant of the resulting quadratic to zero, you will arrive at the same condition $c^2 = a^2m^2 - b^2$. It is a great way to verify your work!

Enter Numerical Value:

Visualized Solution

Visualizing the Hyperbola and Tangent

Standard Hyperbola: $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$
Given Tangent Line: $2 x + y = 1$

Identifying the Directrix

The nearest directrix to the right branch is $x = \frac{a}{e}$ .
The problem states the tangent passes through the intersection of this directrix and the x-axis.

Finding the Intersection Point

Intersection of $x = \frac{a}{e}$ and the x-axis ( $y = 0$ ).
Point $P = (\frac{a}{e}, 0)$ .

Applying the Tangent Condition

The tangent line $2 x + y = 1$ passes through $P (\frac{a}{e}, 0)$ .
Substitute $x = \frac{a}{e}$ and $y = 0$ :
$2 (\frac{a}{e}) + 0 = 1$

Relation Between $a$ and $e$

$2 (\frac{a}{e}) = 1$
Rearranging gives: $a = \frac{e}{2}$

Standard Condition of Tangency

Rewrite tangent as $y = - 2 x + 1$ .
Slope $m = - 2$ , y-intercept $c = 1$ .
Condition for tangency: $c^{2} = a^{2} m^{2} - b^{2}$ .

Substituting $m$ and $c$

Substitute $m = - 2$ and $c = 1$ into the condition:
$1^{2} = a^{2} (- 2)^{2} - b^{2}$
$1 = 4 a^{2} - b^{2}$

Hyperbola Eccentricity Relation

For a hyperbola, $b^{2} = a^{2} (e^{2} - 1)$ .
We need to eliminate $b^{2}$ to find $e$ .

Eliminating $b^{2}$

Substitute $b^{2} = a^{2} (e^{2} - 1)$ into $1 = 4 a^{2} - b^{2}$ :
$1 = 4 a^{2} - a^{2} (e^{2} - 1)$

Simplifying the Equation

$1 = 4 a^{2} - a^{2} e^{2} + a^{2}$
$1 = 5 a^{2} - a^{2} e^{2}$
$1 = a^{2} (5 - e^{2})$

Bringing Back $a = \frac{e}{2}$

Recall from earlier: $a = \frac{e}{2}$ .
Substitute this into $1 = a^{2} (5 - e^{2})$ :
$1 = (\frac{e}{2})^{2} (5 - e^{2})$

Forming the Polynomial

$1 = \frac{e ^{2}}{4} (5 - e^{2})$
Multiply by 4: $4 = 5 e^{2} - e^{4}$
Rearrange: $e^{4} - 5 e^{2} + 4 = 0$

Solving for $e^{2}$

Factorize $e^{4} - 5 e^{2} + 4 = 0$ :
$(e^{2} - 4) (e^{2} - 1) = 0$
Possible values: $e^{2} = 4$ or $e^{2} = 1$

Finalizing the Eccentricity

For a hyperbola, eccentricity $e > 1$ .
Therefore, $e^{2} = 1$ is rejected.
We must have $e^{2} = 4 ⟹ e = 2$ .

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

Solution Diagram

Analyzing the Setup

The hyperbola is defined by the standard equation:

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

We are given a tangent line

2 x + y = 1

that passes through the intersection of the directrix and the

x

-axis. For the right branch of the hyperbola, the directrix is given by the equation

x = \frac{a}{e}

Phase 1

The Geometry of the Directrix

The intersection of the directrix

x = \frac{a}{e}

and the

x

-axis occurs where

y = 0

. Thus, the point of intersection

P

is:

P = (\frac{a}{e}, 0)

Since the line

2 x + y = 1

passes through point

P

, we substitute these coordinates into the line equation:

2 (\frac{a}{e}) + 0 = 1 \Rightarrow a = \frac{e}{2}

Phase 2

The Tangency Condition

The line

2 x + y = 1

can be rewritten in slope-intercept form as

y = - 2 x + 1

. Here, the slope

m = - 2

and the

y

-intercept

c = 1

For a line

y = m x + c

to be tangent to the hyperbola

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

, it must satisfy the condition:

c^{2} = a^{2} m^{2} - b^{2}

Substituting our known values (

m = - 2, c = 1

) into this condition:

1^{2} = a^{2} (- 2)^{2} - b^{2} \Rightarrow 1 = 4 a^{2} - b^{2}

Phase 3

The Algebraic Dance

We utilize the fundamental hyperbola relation

b^{2} = a^{2} (e^{2} - 1)

to eliminate

b^{2}

from our tangency equation:

1 = 4 a^{2} - a^{2} (e^{2} - 1)

Expanding the terms, we obtain:

1 = 4 a^{2} - a^{2} e^{2} + a^{2} \Rightarrow 1 = 5 a^{2} - a^{2} e^{2} \Rightarrow 1 = a^{2} (5 - e^{2})

Now, substitute

a = \frac{e}{2}

into the equation above:

1 = (\frac{e}{2})^{2} (5 - e^{2}) \Rightarrow 1 = \frac{e ^{2}}{4} (5 - e^{2})

Multiplying by

4

yields the following quartic equation:

4 = 5 e^{2} - e^{4} \Rightarrow e^{4} - 5 e^{2} + 4 = 0

Final Calculation

Factoring the quadratic in terms of

e^{2}

(e^{2} - 4) (e^{2} - 1) = 0

This results in two potential values:

e^{2} = 4

e^{2} = 1

. Since the eccentricity of a hyperbola must satisfy

e > 1

, we reject

e^{2} = 1

Therefore,

e^{2} = 4

, which leads to the final result:

$e = 2$

Common Traps & FAQs

Wait, why do we reject

e^{2} = 1

Excellent observation! In the world of conic sections, the eccentricity

e

defines the shape. If

e = 1

, the curve is a parabola. If

0 < e < 1

, it is an ellipse. For a hyperbola, we strictly require

e > 1

. Since our equation gave us

e^{2} = 1

as a root, we must discard it because it violates the fundamental definition of a hyperbola.

Why is the directrix

x = \frac{a}{e}

Think of the directrix as the 'source' of the hyperbola's curvature. For a standard hyperbola

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

, the distance from the center to the directrix is

\frac{a}{e}

. It is a geometric property derived from the focus-directrix definition of conics.

Can I solve this using the discriminant of the quadratic equation?

Yes, you absolutely can! If you substitute

y = 1 - 2 x

into the hyperbola equation and set the discriminant of the resulting quadratic to zero, you will arrive at the same condition

c^{2} = a^{2} m^{2} - b^{2}

. It is a great way to verify your work!

Animated Solution for Mathematics - Conic Sections: The line $2 x + y = 1$ is tangent to the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ . If this line passes through the point of intersection of the nearest directrix and the x-axis, then the eccentricity of the hyperbola is

Enter Numerical Value:

Visualized Solution

Visualizing the Hyperbola and Tangent

Identifying the Directrix

Finding the Intersection Point

Applying the Tangent Condition

Relation Between $a$ and $e$

Standard Condition of Tangency

Substituting $m$ and $c$

Hyperbola Eccentricity Relation

Eliminating $b^{2}$

Simplifying the Equation

Bringing Back $a = \frac{e}{2}$

Forming the Polynomial

Solving for $e^{2}$

Finalizing the Eccentricity

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

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

Visualizing the Hyperbola and Tangent

Identifying the Directrix

Finding the Intersection Point

Applying the Tangent Condition

Relation Between .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e

Standard Condition of Tangency

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

Hyperbola Eccentricity Relation

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

Simplifying the Equation

Bringing Back .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=2e​

Forming the Polynomial

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e2

Finalizing the Eccentricity

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

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the line x−1=0, is a directrix of the hyperbola kx2−y2=6, then the hyperbola passes through the point

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The eccentricity of an ellipse, with its centre at the origin, is 1/2. If one of the directrices is x=4, then the equation of the ellipse is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a directrix of a hyperbola centred at the origin and passing through the point (4,−23​) is 5x=45​ and its eccentricity is e, then :

.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 (3​)kx+ky−43​=0 and 3​x−y−4(3​)k=0 is a conic, whose eccentricity is

Relation Between $a$ and $e$

Substituting $m$ and $c$

Eliminating $b^{2}$

Bringing Back $a = \frac{e}{2}$

Solving for $e^{2}$

If the line $x - 1 = 0$ , is a directrix of the hyperbola $k x^{2} - y^{2} = 6$ , then the hyperbola passes through the point

The eccentricity of an ellipse, with its centre at the origin, is $1/2$ . If one of the directrices is $x = 4$ , then the equation of the ellipse is:

If a directrix of a hyperbola centred at the origin and passing through the point $(4, - 23)$ is $5 x = 45$ and its eccentricity is $e$ , then :

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