Question

describe the key features of a parabola with the equation $x^2 = 40y$. the value of $p$ is $10$. the parabola opens up. the coordinates of the focus are $\boldsymbol{\text{dropdown with (0,0), (0, -10), (0, 10), (10, 0)}}$. the equation for the directrix is $\boldsymbol{\text{box}}$

Explanation:

Step1: Recall the standard form of a parabola

The standard form of a parabola that opens up or down is \(x^{2}=4py\), where the vertex is at \((0,0)\), the focus is at \((0,p)\), and the directrix is \(y = -p\).

Step2: Identify \(p\) from the given equation

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

Step3: Determine the focus coordinates

Since the parabola opens up (as given) and the standard form for a parabola opening up is \(x^{2}=4py\) with focus at \((0,p)\), substituting \(p = 10\), the focus is at \((0,10)\).

Answer:

(0,10)