Question

question graph the parabola given by the equation $x^{2}=-13y$ by first selecting the correct symmetry, then drag the point to one of the endpoints of the latus rectum.

Explanation:

Step1: Identify the form of the parabola

The given equation $x^{2}=- 13y$ is of the form $x^{2} = -4py$. Comparing $x^{2}=-13y$ with $x^{2}=-4py$, we have $4p = 13$, so $p=\frac{13}{4}$.

Step2: Determine the vertex and axis - of - symmetry

For a parabola of the form $x^{2}=-4py$, the vertex is at the origin $(0,0)$. Since the equation involves $x^{2}$, the axis of symmetry is the $y$ - axis (the line $x = 0$).

Step3: Find the endpoints of the latus rectum

The endpoints of the latus rectum of a parabola $x^{2}=-4py$ are given by $(\pm2p,-p)$. Substituting $p = \frac{13}{4}$, we get the endpoints as $(\pm\frac{13}{2},-\frac{13}{4})$.

To graph the parabola:

1. Mark the vertex at the origin $(0,0)$.
2. Since the parabola opens downwards (because the coefficient of $y$ is negative in $x^{2}=-13y$), and the axis of symmetry is $x = 0$.
3. Mark the endpoints of the latus rectum $(\frac{13}{2},-\frac{13}{4})$ and $(-\frac{13}{2},-\frac{13}{4})$. Then draw a smooth curve passing through the vertex and the endpoints of the latus rectum.

Answer:

The axis of symmetry is the $y$ - axis ($x = 0$), and the endpoints of the latus rectum are $(\frac{13}{2},-\frac{13}{4})$ and $(-\frac{13}{2},-\frac{13}{4})$.