The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of i

Q: Why do we reject the negative root when solving for $e$?

Eccentricity is a measure of how 'stretched' the conic section is. For a hyperbola, the foci are further apart than the vertices, meaning $e$ must be greater than $1$. A negative value would be physically meaningless in this geometric context.

Q: Wait, why is the formula $b^2 = a^2(e^2 - 1)$ and not $b^2 = a^2(1 - e^2)$?

That is a classic trap! $b^2 = a^2(1 - e^2)$ is for an ellipse, where $e 1$. We need the term $(e^2 - 1)$ to keep $b^2$ positive.

Q: Can I just divide by $a^2$ without worrying about it being zero?

Yes, absolutely. In the standard hyperbola equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, $a$ represents the semi-transverse axis length. A hyperbola cannot exist if $a = 0$, so $a$ is strictly non-zero. Dividing by $a^2$ is mathematically sound.

Question

The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is :

Options:

(A) $\frac{2}{\sqrt{3}}$

(B) $\sqrt{3}$

(C) $\frac{4}{3}$

(D) $\frac{4}{\sqrt{3}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We begin with the standard equation of a hyperbola:

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

Here, $a$ represents the semi-transverse axis and $b$ represents the semi-conjugate axis. Our goal is to determine the eccentricity $e$ by translating the given geometric properties into algebraic constraints.

### Phase 1: Translating Geometry into Algebra

The problem states that the length of the latus rectum is $8$. Recalling that the length of the latus rectum is defined as $\frac{2b^2}{a}$, we set up the following equation:

$$
\frac{2b^2}{a} = 8
$$

Simplifying this expression, we obtain our first anchor equation:

$$
b^2 = 4a
$$

### Phase 2: The Bridge Between Foci and Conjugate Axis

We are given that the length of the conjugate axis is half the distance between the foci. The length of the conjugate axis is $2b$, and the distance between the foci is $2ae$.

The condition is expressed as:

$$
2b = \frac{1}{2}(2ae) \Rightarrow 2b = ae
$$

To facilitate further calculation, we square both sides to eliminate radicals:

$$
4b^2 = a^2e^2
$$

### Phase 3: The Fundamental Identity

To solve for $e$, we utilize the fundamental identity of a hyperbola, which relates the axes to the eccentricity:

$$
b^2 = a^2(e^2 - 1)
$$

From our second anchor, we know that $b^2 = \frac{a^2e^2}{4}$. Substituting this into the fundamental identity, we get:

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

### Phase 4: The Final Calculation

Since $a 
eq 0$, we can divide both sides of the equation by $a^2$:

$$
\frac{e^2}{4} = e^2 - 1
$$

Multiplying the entire equation by $4$ yields:

$$
e^2 = 4e^2 - 4
$$

Rearranging the terms to solve for $e^2$:

$$
3e^2 = 4 \Rightarrow e^2 = \frac{4}{3}
$$

Taking the square root, we arrive at the final result:

$$
e = \frac{2}{\sqrt{3}}
$$

Since the eccentricity of a hyperbola must satisfy $e > 1$, and $\frac{2}{\sqrt{3}} \approx 1.15$, our result is mathematically consistent. The final eccentricity is **$e = \frac{2}{\sqrt{3}}$**.

Animated Solution for Mathematics - Conic Sections: The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is :

Select Answer:

Visualized Solution

$S t an d a r d H y p er b o l a S e t u p$

$L a t u s R ec t u m C o n d i t i o n$

$S im pl i f y in g f or b^{2}$

$C o nj ug a t e A x i s an d F oc i R e l a t i o n$

$S im pl i f y in g t h e S eco n d C o n d i t i o n$

$T h e F u n d am e n t a l I d e n t i t y$

$S u b s t i t u t in g t h e I d e n t i t y$

$E l imina t in g a^{2}$

$S o l v in g f or e^{2}$

$F ina l V a l u eo f E cce n t r i c i t y$

The Sigma Insight: Foci, Directrices, and Eccentricity

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Conic Sections: The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is :

Select Answer:

Visualized Solution

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Eliminatinga2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solvingfore2

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