find the focus, directrix, axis of symmetry, focal diameter, and endpoints of latus rectum of the parabola

Given the equation of the parabola:

\[
x = -\frac{1}{5}y^2
\]

Let's rewrite this in the standard form to find the required properties. The equation can be rearranged as:

\[
y^2 = -5x
\]

This matches the standard form of a parabola that opens horizontally (either left or right):

\[
y^2 = 4px
\]

### 1. **Focus:**
The general formula for the focus of the parabola \( y^2 = 4px \) is \( \left(p, 0\right) \).

In our case, \( 4p = -5 \), so \( p = -\frac{5}{4} \).

Thus, the focus is at:

\[
\left(-\frac{5}{4}, 0\right)
\]

### 2. **Directrix:**
The directrix of a parabola \( y^2 = 4px \) is a vertical line given by \( x = -p \).

For our parabola:

\[
x = \frac{5}{4}
\]

### 3. **Axis of Symmetry:**
The axis of symmetry is the line along which the parabola is symmetric. For a horizontally oriented parabola like this, the axis of symmetry is the horizontal line through the vertex:

\[
y = 0
\]

### 4. **Focal Diameter:**
The focal diameter (also known as the latus rectum) is the distance between the endpoints of the latus rectum. The length of the focal diameter is \( |4p| \).

In our case:

\[
|4p| = 5
\]

### 5. **Endpoints of the Latus Rectum:**
The latus rectum is a line segment perpendicular to the axis of symmetry that passes through the focus. Its endpoints can be found using the coordinates of the focus \( \left(-\frac{5}{4}, 0\right) \).

The endpoints of the latus rectum are:

\[
\left(-\frac{5}{4}, \pm \frac{|4p|}{2}\right)
\]

So, substituting \( |4p| = 5 \):

\[
\left(-\frac{5}{4}, \pm \frac{5}{2}\right)
\]

Which gives the endpoints as:

\[
\left(-\frac{5}{4}, \frac{5}{2}\right) \quad \text{and} \quad \left(-\frac{5}{4}, -\frac{5}{2}\right)
\]

### Summary:
- **Focus**: \( \left(-\frac{5}{4}, 0\right) \)
- **Directrix**: \( x = \frac{5}{4} \)
- **Axis of Symmetry**: \( y = 0 \)
- **Focal Diameter**: \( 5 \)
- **Endpoints of Latus Rectum**: \( \left(-\frac{5}{4}, \frac{5}{2}\right) \) and \( \left(-\frac{5}{4}, -\frac{5}{2}\right) \)

Would you like more details or have any questions?

Here are 5 related questions:
1. How does the value of \( p \) affect the position of the focus and directrix?
2. What are the latus rectum and its significance in a parabola?
3. How would the focus and directrix change if the parabola opened upward instead of leftward?
4. Can you find the equation of the parabola if the focus is given?
5. How can you derive the equation of the directrix from the parabola's equation?

**Tip**: Remember that the distance from any point on the parabola to the focus is equal to its perpendicular distance to the directrix. This property can help in solving problems related to parabolas.

Learn how to find the focus, directrix, axis of symmetry, focal diameter, and endpoints of latus rectum of a parabola with a detailed step-by-step solution. This problem involves understanding parabolas, including their properties and geometric relationships. Suitable for high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation