Question

describe the key features of a parabola with the equation ( x^{2}=40y ).
the value of ( p ) is
the parabola opens
the coordinates of the focus are
the equation for the directrix is

Explanation:

Step1: Recall the standard form of parabola

The standard form of a parabola that opens up or down is \(x^{2}=4py\).

Step2: Compare with given equation

Given \(x^{2}=40y\), comparing with \(x^{2}=4py\), we have \(4p = 40\), so \(p=\frac{40}{4}=10\).

Step3: Determine the direction of opening

Since the coefficient of \(y\) is positive (\(x^{2}=40y\) where \(4p = 40>0\)), the parabola opens up.

Step4: Find the focus coordinates

For the parabola \(x^{2}=4py\), the focus is at \((0,p)\). Substituting \(p = 10\), the focus is \((0,10)\).

Step5: Find the directrix equation

For the parabola \(x^{2}=4py\), the directrix is \(y=-p\). Substituting \(p = 10\), the directrix is \(y=- 10\).

Answer:

The value of \(p\) is \(10\).
The parabola opens up.
The coordinates of the focus are \((0,10)\).
The equation for the directrix is \(y = - 10\).