Let the foci of a hyperbola be (1, 14) and (1, -12). If it passes through the point (1, 6), then the

Q: Wait, why is $(1, 6)$ a vertex? Can't it just be a random point on the curve?

Great question! A hyperbola is symmetric about its transverse axis. If a point on the hyperbola has the same $x$-coordinate as the foci, it must lie on the transverse axis. The only points on the hyperbola that also lie on the transverse axis are the vertices. Therefore, $(1, 6)$ must be a vertex.

Q: Why is the distance between foci $2ae$?

This is a fundamental property of the hyperbola. By definition, the distance from the center to a focus is $ae$. Since there are two foci, one on each side of the center, the total distance is $ae + ae = 2ae$.

Q: Can I solve this using the standard equation $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$?

Absolutely! You could plug in the center $(1, 1)$ and the vertex $(1, 6)$ to find $a$, and then use the point $(1, 6)$... wait, that doesn't help find $b$. You would need another point or the eccentricity. The method we used—leveraging the geometric properties—is much faster and less prone to algebraic errors.

Question

Let the foci of a hyperbola be $(1,14)$ and $(1,-12)$. If it passes through the point $(1,6),$ then the length of its latus-rectum is :

Options:

(A) $\frac{24}{5}$

(B) $\frac{25}{6}$

(C) $\frac{144}{5}$

(D) $\frac{288}{5}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane. You are given two points, $S(1, 14)$ and $S'(1, -12)$, which are the foci of a hyperbola.

The first thing a master of conics does is look for symmetry. Notice that both points share the exact same $x$-coordinate, $x=1$.

This is our anchor! It tells us that the transverse axis—the line passing through the foci—is a vertical line, $x=1$. This simple observation simplifies our entire journey.

### The Center of the Universe

Every hyperbola has a center, a point of perfect symmetry. Geometrically, this center $C$ must lie exactly midway between the two foci.

We use the midpoint formula: 
$$C = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} ight)$$

Substituting our values, we get:
$$C = \left( 1, \frac{14 + (-12)}{2} ight) = (1, 1)$$

This point $(1, 1)$ is the heart of our hyperbola.

### The "Aha!" Moment

Now, consider the point $(1, 6)$ through which our hyperbola passes. Look at its $x$-coordinate—it is $1$.

This means the point lies directly on our transverse axis. In the world of hyperbolas, any point that lies on the curve and the transverse axis is, by definition, a vertex.

So, $V(1, 6)$ is our vertex. The distance from the center $C(1, 1)$ to the vertex $V(1, 6)$ is the semi-transverse axis, $a$. Thus, $a = |6 - 1| = 5$. We have unlocked the first key parameter!

### The Algebra of the Hyperbola

Next, we need the distance between the foci. By definition, this distance is $2ae$.

Looking at our coordinates, the distance is $|14 - (-12)| = 26$. Therefore, $2ae = 26$, which means $ae = 13$.

Now, we have $a=5$ and $ae=13$. We need the semi-conjugate axis $b$ to find the latus rectum. We use the fundamental relation:
$$b^2 = a^2(e^2 - 1)$$

Expanding this, we get:
$$b^2 = a^2e^2 - a^2 = (ae)^2 - a^2$$

Substituting our values:
$$b^2 = (13)^2 - (5)^2 = 169 - 25 = 144$$

### The Final Victory

The length of the latus rectum is given by the elegant formula:
$$	ext{Length} = \frac{2b^2}{a}$$

We have everything we need: $b^2 = 144$ and $a = 5$. Plugging these in, we get:
$$\frac{2(144)}{5} = \frac{288}{5}$$

And there it is! The math isn't just about numbers; it's about the beautiful, rigid structure of the hyperbola revealing itself through simple geometric properties.

The final length of the latus rectum is **$\frac{288}{5}$** (or **$57.6$**).

Animated Solution for Mathematics - Conic Sections: Let the foci of a hyperbola be $(1, 14)$ and $(1, - 12)$ . If it passes through the point $(1, 6),$ then the length of its latus-rectum is :

Select Answer:

Visualized Solution

Locate the Foci

Find the Center $C$

Calculate Center Coordinates

Distance Between Foci ( $2 a e$ )

Calculate $a e$

Identify the Vertex $V$

Calculate Semi-axis $a$

Calculate $b^{2}$

Evaluate $b^{2}$

Length of Latus Rectum

Final Answer

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

Analyzing the Setup

The Center of the Universe

The "Aha!" Moment

The Algebra of the Hyperbola

The Final Victory

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Visualized Solution

Locate the Foci

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

Calculate Center Coordinates

Distance Between Foci ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2ae)

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

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

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

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

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