Let the eccentricity of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be a reciprocal to that of the ellipse $x^2 + 4y^2 = 4$. If the hyperbola passes through a focus of the ellipse, then Options: (A) the equation of the hyperbola is $\frac{x^2}{3} - \frac{y^2}{2} = 1$ (B) a focus of the hyperbola is $(2, 0)$ (C) the eccentricity of the hyperbola is $\sqrt{\frac{5}{3}}$ (D) the equation of the hyperbola is $x^2 - 3y^2 = 3$

### Analyzing the Ellipse We begin with the ellipse defined by the equation $x^2 + 4y^2 = 4$. To reveal its geometric properties, we convert this into the standard form by dividing the entire equation by $4$: $$\frac{x^2}{4} + \frac{y^2}{1} = 1$$ From this standard form, we identify the parameters $a_e^2 = 4$ and $b_e^2 = 1$. We then calculate the eccentricity $e_e$ of the ellipse: $$e_e = \sqrt{1 - \frac{b_e^2}{a_e^2}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$$ This value, $e_e = \frac{\sqrt{3}}{2}$, serves as the fundamental "DNA" of our ellipse. ### The Bridge of Eccentricity The problem states that the eccentricity of the hyperbola, $e_h$, is the reciprocal of the ellipse's eccentricity. Therefore, we calculate: $$e_h = \frac{1}{e_e} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}$$ Next, we locate the foci of the ellipse using the coordinates $(\pm a_e e_e, 0)$. Substituting our known values, we find: $$(\pm 2 \cdot \frac{\sqrt{3}}{2}, 0) = (\pm \sqrt{3}, 0)$$ We take the positive focus, $(\sqrt{3}, 0)$, and require that the hyperbola passes through this point. Substituting this into the standard hyperbola equation $\frac{x^2}{a_h^2} - \frac{y^2}{b_h^2} = 1$, we obtain: $$\frac{(\sqrt{3})^2}{a_h^2} - \frac{0^2}{b_h^2} = 1 \Rightarrow \frac{3}{a_h^2} = 1 \Rightarrow a_h^2 = 3$$ ### The Final Synthesis With $a_h^2 = 3$ and $e_h = \frac{2}{\sqrt{3}}$, we determine $b_h^2$ using the hyperbola identity $b_h^2 = a_h^2(e_h^2 - 1)$: $$b_h^2 = 3 \left( \left( \frac{2}{\sqrt{3}} \right)^2 - 1 \right) = 3 \left( \frac{4}{3} - 1 \right) = 3 \left( \frac{1}{3} \right) = 1$$ Substituting $a_h^2 = 3$ and $b_h^2 = 1$ back into the standard form, the equation of the hyperbola is: $$\frac{x^2}{3} - \frac{y^2}{1} = 1$$ This simplifies to the final result: **$x^2 - 3y^2 = 3$**. To verify, we calculate the foci of the hyperbola using $(\pm a_h e_h, 0)$, which yields $(\pm \sqrt{3} \cdot \frac{2}{\sqrt{3}}, 0) = (\pm 2, 0)$. The geometry is consistent and complete.

JEE Advanced 2011

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: Let the eccentricity of the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ be a reciprocal to that of the ellipse $x^{2} + 4 y^{2} = 4$ . If the hyperbola passes through a focus of the ellipse, then

Select Answer:

* Multiple Correct

Visualized Solution

Given Ellipse Equation

Given Ellipse: $x^{2} + 4 y^{2} = 4$
Standard form: $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$

Standard Form of Ellipse

Divide by $4$ : $\frac{x ^{2}}{4} + \frac{4 y ^{2}}{4} = \frac{4}{4}$
$\frac{x ^{2}}{4} + \frac{y ^{2}}{1} = 1$
Comparing gives $a_{e}^{2} = 4$ and $b_{e}^{2} = 1$

Ellipse Eccentricity $e_{e}$

Eccentricity formula: $e_{e} = 1 - \frac{b _{e}^{2}}{a _{e}^{2}}$
$e_{e} = 1 - \frac{1}{4} = \frac{3}{4}$
$e_{e} = \frac{3}{2}$

Locating Ellipse Foci

Foci of ellipse: $(\pm a_{e} e_{e}, 0)$
Since $a_{e} = 2$ and $e_{e} = \frac{3}{2}$
Foci: $(\pm 2 \cdot \frac{3}{2}, 0) = (\pm 3, 0)$

Hyperbola Eccentricity $e_{h}$

Given: $e_{h} = \frac{1}{e _{e}}$
$e_{h} = \frac{1}{\frac{3}{2}}$
$e_{h} = \frac{2}{3}$

Hyperbola Vertex

Hyperbola passes through a focus of the ellipse.
Let's use the positive focus: $(3, 0)$
Substitute in $\frac{x ^{2}}{a _{h}^{2}} - \frac{y ^{2}}{b _{h}^{2}} = 1$

Finding $a_{h}^{2}$

$\frac{( 3 ) ^{2}}{a _{h}^{2}} - \frac{0 ^{2}}{b _{h}^{2}} = 1$
$\frac{3}{a _{h}^{2}} = 1$
$a_{h}^{2} = 3$

Calculating $b_{h}^{2}$

Relation: $b_{h}^{2} = a_{h}^{2} (e_{h}^{2} - 1)$
$b_{h}^{2} = 3 ((\frac{2}{3})^{2} - 1)$
$b_{h}^{2} = 3 (\frac{4}{3} - 1)$

Finalizing $b_{h}^{2}$

$b_{h}^{2} = 3 (\frac{4 - 3}{3})$
$b_{h}^{2} = 3 \cdot \frac{1}{3}$
$b_{h}^{2} = 1$

Final Hyperbola Equation

Equation: $\frac{x ^{2}}{3} - \frac{y ^{2}}{1} = 1$
Multiply by $3$ : $x^{2} - 3 y^{2} = 3$
Option 4 is correct.

Finding Hyperbola Focus

Focus of hyperbola: $(\pm a_{h} e_{h}, 0)$
We know $a_{h} = 3$ and $e_{h} = \frac{2}{3}$
Focus: $(\pm 3 \cdot \frac{2}{3}, 0)$

Verifying Option 2

Focus: $(\pm 2, 0)$
One of the foci is $(2, 0)$
Option 2 is also correct.

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The Sigma Insight: Foci, Directrices, and Eccentricity

Solution Diagram

Analyzing the Ellipse

We begin with the ellipse defined by the equation

x^{2} + 4 y^{2} = 4

. To reveal its geometric properties, we convert this into the standard form by dividing the entire equation by

4

\frac{x ^{2}}{4} + \frac{y ^{2}}{1} = 1

From this standard form, we identify the parameters

a_{e}^{2} = 4

and

b_{e}^{2} = 1

. We then calculate the eccentricity

e_{e}

of the ellipse:

e_{e} = 1 - \frac{b _{e}^{2}}{a _{e}^{2}} = 1 - \frac{1}{4} = \frac{3}{2}

This value,

e_{e} = \frac{3}{2}

, serves as the fundamental "DNA" of our ellipse.

The Bridge of Eccentricity

The problem states that the eccentricity of the hyperbola,

e_{h}

, is the reciprocal of the ellipse's eccentricity. Therefore, we calculate:

e_{h} = \frac{1}{e _{e}} = \frac{1}{\frac{3}{2}} = \frac{2}{3}

Next, we locate the foci of the ellipse using the coordinates

(\pm a_{e} e_{e}, 0)

. Substituting our known values, we find:

(\pm 2 \cdot \frac{3}{2}, 0) = (\pm 3, 0)

We take the positive focus,

(3, 0)

, and require that the hyperbola passes through this point. Substituting this into the standard hyperbola equation

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

, we obtain:

\frac{( 3 ) ^{2}}{a _{h}^{2}} - \frac{0 ^{2}}{b _{h}^{2}} = 1 \Rightarrow \frac{3}{a _{h}^{2}} = 1 \Rightarrow a_{h}^{2} = 3

The Final Synthesis

With

a_{h}^{2} = 3

and

e_{h} = \frac{2}{3}

, we determine

b_{h}^{2}

using the hyperbola identity

b_{h}^{2} = a_{h}^{2} (e_{h}^{2} - 1)

b_{h}^{2} = 3 ((\frac{2}{3})^{2} - 1) = 3 (\frac{4}{3} - 1) = 3 (\frac{1}{3}) = 1

Substituting

a_{h}^{2} = 3

and

b_{h}^{2} = 1

back into the standard form, the equation of the hyperbola is:

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

This simplifies to the final result: $x^{2} - 3 y^{2} = 3$ .

To verify, we calculate the foci of the hyperbola using

(\pm a_{h} e_{h}, 0)

, which yields

(\pm 3 \cdot \frac{2}{3}, 0) = (\pm 2, 0)

. The geometry is consistent and complete.

Common Traps & FAQs

Wait, why do we use the formula

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

for the hyperbola but

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

for the ellipse? It's confusing!

That is a fantastic observation! Think of it as the 'nature' of the curves. An ellipse is a bounded, closed loop, so its eccentricity

e

must be less than

1

, making

(1 - e^{2})

positive. A hyperbola is an unbounded, open curve, so its eccentricity

e

must be greater than

1

, making

(e^{2} - 1)

positive. The formula is simply ensuring that

b^{2}

remains a positive real number. It is all about maintaining the physical reality of the geometry!

Why does the hyperbola pass through the focus of the ellipse? Is there a special geometric property here?

In this specific problem, it is a given constraint, not necessarily a universal property of all ellipses and hyperbolas. It acts as a 'bridge' condition that forces the two curves to intersect at a specific coordinate. In JEE problems, always look for these 'bridge' constraints—they are the keys that unlock the algebraic path forward.

How do I know if I should use the positive or negative focus of the ellipse?

Because both the ellipse and the hyperbola are symmetric about the coordinate axes, the hyperbola will pass through both

(3, 0)

and

(- 3, 0)

. Using either one will yield the exact same result for

a_{h}^{2}

. Symmetry is your best friend in coordinate geometry—use it to simplify your life!

Animated Solution for Mathematics - Conic Sections: Let the eccentricity of the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ be a reciprocal to that of the ellipse $x^{2} + 4 y^{2} = 4$ . If the hyperbola passes through a focus of the ellipse, then

Select Answer:

Visualized Solution

Given Ellipse Equation

Standard Form of Ellipse

Ellipse Eccentricity $e_{e}$

Locating Ellipse Foci

Hyperbola Eccentricity $e_{h}$

Hyperbola Vertex

Finding $a_{h}^{2}$

Calculating $b_{h}^{2}$

Finalizing $b_{h}^{2}$

Final Hyperbola Equation

Finding Hyperbola Focus

Verifying Option 2

The Sigma Insight: Foci, Directrices, and Eccentricity

Analyzing the Ellipse

The Bridge of Eccentricity

The Final Synthesis

Similar Questions

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Select Answer:

Visualized Solution

Given Ellipse Equation

Standard Form of Ellipse

Ellipse Eccentricity .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ee​

Locating Ellipse Foci

Hyperbola Eccentricity .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } eh​

Hyperbola Vertex

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ah2​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } bh2​

Finalizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } bh2​

Final Hyperbola Equation

Finding Hyperbola Focus

Verifying Option 2

The Sigma Insight: Foci, Directrices, and Eccentricity

Analyzing the Ellipse

The Bridge of Eccentricity

The Final Synthesis

Similar Questions

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Hyperbola Eccentricity $e_{h}$

Finding $a_{h}^{2}$

Calculating $b_{h}^{2}$

Finalizing $b_{h}^{2}$

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