Consider a hyperbola H having centre at the origin and foci on the x-axis. Let C1 be the circle touc

Q: Why does the circle $C_1$ touch the hyperbola at the vertex? Can't it touch somewhere else?

Great question! Because the hyperbola is symmetric about both the $x$ and $y$ axes, and the circle is centered at the origin, the points of contact must respect this symmetry. The closest points on the hyperbola to the origin are the vertices $(\pm a, 0)$. Any circle centered at the origin that expands to touch the hyperbola will hit these vertices first.

Q: I'm confused about the radius of $C_2$. Why is it $ae - a$?

Think about the geometry. The center of $C_2$ is at the focus $(ae, 0)$. The circle touches the hyperbola at the vertex $(a, 0)$. The distance between these two points on the $x$-axis is simply the difference of their coordinates: $ae - a$. That distance is the radius of the circle.

Q: When solving $(e-1)^2 = 1/9$, I got $e-1 = -1/3$ as well. Why do we ignore it?

In the world of hyperbolas, the eccentricity $e$ is defined as being strictly greater than $1$. If we took $e-1 = -1/3$, we would get $e = 2/3$, which is less than $1$. That would describe an ellipse, not a hyperbola! Always check your constraints.

Question

Consider a hyperbola $H$ having centre at the origin and foci on the $x$-axis. Let $C_1$ be the circle touching the hyperbola $H$ and having the centre at the origin. Let $C_2$ be the circle touching the hyperbola $H$ at its vertex and having the centre at one of its foci. If areas (in sq units) of $C_1$ and $C_2$ are $36\pi$ and $4\pi$, respectively, then the length (in units) of latus rectum of $H$ is

Options:

(A) $\frac{14}{3}$

(B) $\frac{28}{3}$

(C) $\frac{11}{3}$

(D) $\frac{10}{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are exploring the elegant architecture of the hyperbola. Imagine you are standing at the origin of a coordinate plane.

Before you, a hyperbola $H$ stretches out, its branches opening wide, centered perfectly at your feet. Its foci lie along the $x$-axis, acting as the gravitational anchors of this beautiful curve. The standard equation is given by:

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

### The Guardian Circle $C_1$

We are introduced to a circle $C_1$, also centered at the origin. Because the hyperbola is symmetric and the circle is centered at the origin, the circle must touch the hyperbola at its closest points—the vertices $(\pm a, 0)$.

This implies the radius of $C_1$ is simply the distance from the origin to the vertex, which is $a$. We are given the area of $C_1$ as $36\pi$. Using the area formula $\pi r_1^2 = 36\pi$, we find:

$$a^2 = 36 \Rightarrow a = 6$$

### The Focus-Anchored Circle $C_2$

Now, we consider a second circle, $C_2$. Its center is at one of the foci, $(ae, 0)$, and it touches the hyperbola at the vertex $(a, 0)$.

The radius $r_2$ of this circle is the distance between the focus and the vertex. Mathematically, this is expressed as:

$$r_2 = ae - a = a(e - 1)$$

We are told the area of $C_2$ is $4\pi$. Applying the area formula $\pi r_2^2 = 4\pi$, we get $r_2^2 = 4$, so $r_2 = 2$. Substituting our value for $a$:

$$6(e - 1) = 2 \Rightarrow e - 1 = \frac{1}{3} \Rightarrow e = \frac{4}{3}$$

### The Algebraic Synthesis

To find the latus rectum, we need the parameter $b^2$. We reach into our toolkit for the fundamental identity of the hyperbola:

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

Substituting our known values:

$$b^2 = 36 \left( \left( \frac{4}{3} ight)^2 - 1 ight) = 36 \left( \frac{16}{9} - 1 ight) = 36 \left( \frac{7}{9} ight)$$

The calculation yields $b^2 = 4 	imes 7 = 28$.

### The Final Victory

The length of the latus rectum $L$ is defined by the formula:

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

Substituting $b^2 = 28$ and $a = 6$, we obtain:

$$L = \frac{2(28)}{6} = \frac{28}{3}$$

The final length of the latus rectum is **$\frac{28}{3}$**. You have successfully navigated the geometry and algebraic identities to arrive at this precise result.

Select Answer:

Visualized Solution

$S t an d a r d E q u a t i o n o f H y p er b o l a$

$C i r c l e C_{1} T o u c hin g H y p er b o l a$

$A r e a o f C i r c l e C_{1}$

$F in d in g S e mi - M aj or A x i s a$

$C i r c l e C_{2} a tF oc u s$

$R a d i u so f C i r c l e C_{2}$

$A r e a o f C i r c l e C_{2}$

$S im pl i f y in g t h e E q u a t i o n$

$F in d in g E cce n t r i c i t y e$

$H y p er b o l a I d e n t i t y f or b^{2}$

$C a l c u l a t in g b^{2}$

$F or m u l a f or L a t u s R ec t u m$

$F ina l L e n g t h o f L a t u s R ec t u m$

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

Analyzing the Setup

The Guardian Circle $C_{1}$

The Focus-Anchored Circle $C_{2}$

The Algebraic Synthesis

The Final Victory

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CircleC1​TouchingHyperbola

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AreaofCircleC1​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CircleC2​atFocus

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } RadiusofCircleC2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AreaofCircleC2​

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } HyperbolaIdentityforb2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculatingb2

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

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