Key Concept: Definition and Properties of a Hyperbola

A hyperbola is defined as the locus of all points in a plane such that the absolute difference of the distances from two fixed points (called foci) is a constant. This constant is equal to $2a$, where $a$ is the length of the semi-transverse axis. Mathematically, for any point $P$ on the hyperbola and foci $F_1, F_2$, we have: $|PF_1 - PF_2| = 2a$ The length of the latus-rectum (LR) of a hyperbola is given by the formula: $LR = \frac{2b^2}{a}$ where $b$ is the length of the semi-conjugate axis. The relationship between $a$, $b$, and the semi-focal distance $c$ (distance from the center to each focus) for a hyperbola is $c^2 = a^2 + b^2$.

Our goal is to find the length of the latus-rectum, which means we need to determine the values of $a$ and $b^2$.

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Step-by-Step Solution

1. Identify Given Information and the Objective

We are provided with:

• Foci of the hyperbola: $F_1 = (1,14)$ and $F_2 = (1,-12)$.
• A point on the hyperbola: $P = (1,6)$.

Our objective is to calculate the length of the latus-rectum, $LR = \frac{2b^2}{a}$. To do this, we must first find $a$ and $b$.

2. Leverage the Definition of a Hyperbola to Find 'a'

• Why this step? The defining property of a hyperbola directly connects the distances from a point on the hyperbola to its foci with the length of the transverse axis ($2a$). This is the most direct method to find $a$ when a point on the curve and the foci are given.

First, we calculate the distance from the given point $P(1,6)$ to each focus $F_1(1,14)$ and $F_2(1,-12)$ using the distance formula $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.

• Distance $PF_1$: $PF_1 = \sqrt{(1-1)^2 + (6-14)^2} = \sqrt{0^2 + (-8)^2} = \sqrt{64} = 8$

• Distance $PF_2$: $PF_2 = \sqrt{(1-1)^2 + (6-(-12))^2} = \sqrt{0^2 + (6+12)^2} = \sqrt{0^2 + (18)^2} = \sqrt{324} = 18$

Now, we apply the definition of the hyperbola: $|PF_1 - PF_2| = 2a$ $|8 - 18| = 2a$ $|-10| = 2a$ $10 = 2a$ $a = 5$ Thus, the length of the semi-transverse axis is $a=5$.

Tip: Notice that the x-coordinates of the foci ($F_1$ and $F_2$) are both $1$. This indicates that the transverse axis of the hyperbola is a vertical line ($x=1$). The point $P(1,6)$ also has an x-coordinate of $1$, which means it lies on the transverse axis. In fact, since $(1,6)$ lies between $(1,-12)$ and $(1,14)$, and the center is $(1,1)$ (as calculated below), the point $(1,6)$ is one of the vertices of the hyperbola. If $P$ is a vertex, then $PF_1 = a+c$ and $PF_2 = c-a$ (or vice versa), and $|(a+c)-(c-a)| = |2a| = 2a$. This is consistent with our calculation.

3. Determine the Center and Semi-Focal Distance 'c'

• Why this step? The center of the hyperbola is the midpoint of the segment connecting the two foci. The distance between the foci is $2c$, and finding $c$ is crucial because it relates $a$ and $b$ through the fundamental equation $c^2 = a^2 + b^2$.

The center of the hyperbola, $C(h,k)$, is the midpoint of $F_1(1,14)$ and $F_2(1,-12)$: $C = \left(\frac{1+1}{2}, \frac{14+(-12)}{2}\right) = \left(\frac{2}{2}, \frac{2}{2}\right) = (1,1)$ The distance between the foci $F_1(1,14)$ and $F_2(1,-12)$ is $2c$. $2c = \sqrt{(1-1)^2 + (14 - (-12))^2} = \sqrt{0^2 + (14+12)^2} = \sqrt{0^2 + 26^2} = \sqrt{676} = 26$ $2c = 26 \implies c = 13$ So, the semi-focal distance is $c=13$.

4. Calculate the Length of the Semi-Conjugate Axis 'b'

• Why this step? We need $b^2$ for the latus-rectum formula. The relationship $c^2 = a^2 + b^2$ is the fundamental equation connecting $a$, $b$, and $c$ for a hyperbola.

For a hyperbola, the relationship between $a$, $b$, and $c$ is given by: $c^2 = a^2 + b^2$ We have already found $a=5$ and $c=13$. Substitute these values into the equation: $(13)^2 = (5)^2 + b^2$ $169 = 25 + b^2$ $b^2 = 169 - 25$ $b^2 = 144$

Common Mistake Alert: It's crucial to remember the correct relationship for a hyperbola: $c^2 = a^2 + b^2$. For an ellipse, the relationship is $a^2 = b^2 + c^2$ (where $a$ is the semi-major axis, which is always the largest). In a hyperbola, $c$ is always the largest of $a,b,c$ (because the foci are outside the vertices), so $c^2$ must be the sum of the other two squares. Do not confuse these two formulas!

5. Calculate the Length of the Latus-Rectum

• Why this step? This is the final calculation to answer the question, using the values of $a$ and $b^2$ that we have determined.

The formula for the length of the latus-rectum of a hyperbola is: $LR = \frac{2b^2}{a}$ Substitute the values $a=5$ and $b^2=144$: $LR = \frac{2 \times 144}{5}$ $LR = \frac{288}{5}$

6. Compare with Options

The calculated length of the latus-rectum is $\frac{288}{5}$. Let's check the given options: (A) $\frac{25}{6}$ (B) $\frac{144}{5}$ (C) $\frac{288}{5}$ (D) $\frac{24}{5}$

Our result matches option (C).

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Summary and Key Takeaway

This problem is an excellent test of your understanding of the fundamental properties of a hyperbola:

1. Definition of a Hyperbola: The absolute difference of distances from any point on the hyperbola to its foci is $2a$. This is the primary tool to find $a$.
2. Foci and Center: The center of the hyperbola is the midpoint of its foci, and the distance between the foci is $2c$.
3. Fundamental Relationship: For a hyperbola, $c^2 = a^2 + b^2$. Remember this distinction from the ellipse relationship.
4. Latus Rectum Formula: $LR = \frac{2b^2}{a}$.

Always begin by utilizing the definition of the conic section when a point on the curve is given along with the foci. Systematically determine $a$, $c$, and then $b$ (or $b^2$) to arrive at the final answer.