Semi Latus Rectum of Hyperbola given Linear Eccentricity and Semi Conjugate Axis Calculator

Formula Used:

\[ L_{Semi} = \frac{\sqrt{\frac{(2 \cdot b^2)^2}{c^2 - b^2}}}{2} \]

Semi Latus Rectum of Hyperbola (L_Semi):

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1. What is Semi Latus Rectum of Hyperbola?

The Semi Latus Rectum of Hyperbola is half of the line segment passing through any of the foci and perpendicular to the transverse axis whose ends are on the Hyperbola. It is an important parameter in describing the geometry of a hyperbola.

2. How Does the Calculator Work?

The calculator uses the formula:

\[ L_{Semi} = \frac{\sqrt{\frac{(2 \cdot b^2)^2}{c^2 - b^2}}}{2} \]

Where:

\( L_{Semi} \) — Semi Latus Rectum of Hyperbola (m)
\( b \) — Semi Conjugate Axis of Hyperbola (m)
\( c \) — Linear Eccentricity of Hyperbola (m)

Explanation: This formula calculates the semi latus rectum based on the semi conjugate axis and linear eccentricity of the hyperbola.

3. Importance of Semi Latus Rectum Calculation

Details: The semi latus rectum is crucial in understanding the focal properties and overall shape of a hyperbola. It helps in various applications including orbital mechanics and optical systems.

4. Using the Calculator

Tips: Enter the semi conjugate axis (b) and linear eccentricity (c) in meters. Ensure c > b for valid hyperbola parameters. All values must be positive.

5. Frequently Asked Questions (FAQ)

Q1: What is the relationship between semi latus rectum and other hyperbola parameters?
A: The semi latus rectum is related to the semi transverse axis (a) and semi conjugate axis (b) through the formula \( L_{Semi} = \frac{b^2}{a} \).

Q2: How does linear eccentricity relate to the hyperbola's foci?
A: Linear eccentricity (c) is half the distance between the two foci of the hyperbola, and is related to a and b by \( c^2 = a^2 + b^2 \).

Q3: Can the semi latus rectum be negative?
A: No, the semi latus rectum is always a positive value as it represents a physical length.

Q4: What are typical units for these measurements?
A: While meters are used here, any consistent length unit can be used (cm, mm, etc.) as long as all inputs use the same unit.

Q5: When is this calculation most useful?
A: This calculation is particularly useful in physics and engineering applications involving hyperbolic trajectories, such as in orbital mechanics or antenna design.