The end points of the latus rectum of the parabola `x ^(2) + 5y =0` is

Options

A. `(pm (5)/(2), -(5)/(4))`
B. `(pm (2)/(5), pm (4)/(5))`
C. `(pm (4)/(5), pm (4)/(5))`
D. `(pm (5)/(4), -(5)/(2))`

Answer

To find the endpoints of the latus rectum of the parabola given by the equation \( x^2 + 5y = 0 \), we can follow these steps:

### Step 1: Rewrite the equation in standard form
The given equation is:
\[ x^2 + 5y = 0 \]
We can rearrange this to express \( y \) in terms of \( x \):
\[ 5y = -x^2 \implies y = -\frac{1}{5}x^2 \]
This is a parabola that opens downwards.

### Step 2: Identify the value of \( a \)
The standard form of a parabola that opens downwards is given by:
\[ x^2 = -4ay \]
From our rearranged equation \( y = -\frac{1}{5}x^2 \), we can see that:
\[ 4a = \frac{1}{5} \implies a = \frac{1}{20} \]

### Step 3: Determine the coordinates of the focus
The focus of the parabola is located at the point \( (0, -a) \):
\[ \text{Focus} = \left(0, -\frac{1}{20}\right) \]

### Step 4: Calculate the endpoints of the latus rectum
The endpoints of the latus rectum are given by the points:
\[ \left(-2a, -a\right) \quad \text{and} \quad \left(2a, -a\right) \]
Substituting the value of \( a \):
\[ -2a = -2 \times \frac{1}{20} = -\frac{1}{10} \]
\[ 2a = 2 \times \frac{1}{20} = \frac{1}{10} \]
Thus, the coordinates of the endpoints of the latus rectum are:
\[ \left(-\frac{1}{10}, -\frac{1}{20}\right) \quad \text{and} \quad \left(\frac{1}{10}, -\frac{1}{20}\right) \]

### Final Answer
The endpoints of the latus rectum of the parabola \( x^2 + 5y = 0 \) are:
\[ \left(-\frac{1}{10}, -\frac{1}{20}\right) \quad \text{and} \quad \left(\frac{1}{10}, -\frac{1}{20}\right) \]

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