Question

1. vertex at origin, opens right,

\frac{1}{8} units between the vertex and focus

1. vertex: (4, 2), axis of symmetry: x = 4,

length of latus rectum = \frac{1}{3}, a > 0

Explanation:

Problem 20:

Step1: Recall parabola standard form

For a parabola with vertex at origin \((0,0)\) and opening right, the standard form is \(y^2 = 4px\), where \(p\) is the distance from vertex to focus.

Step2: Determine \(p\) value

Given the distance between vertex and focus is \(\frac{1}{8}\), so \(p=\frac{1}{8}\).

Step3: Substitute \(p\) into equation

Substitute \(p = \frac{1}{8}\) into \(y^2=4px\):
\(y^2 = 4\times\frac{1}{8}x\)
Simplify: \(y^2=\frac{1}{2}x\)

Step1: Recall parabola standard form

Since vertex is \((4,2)\) and axis of symmetry is \(x = 4\) (vertical line? No, \(x=4\) is vertical, but wait, axis of symmetry \(x = 4\) means the parabola opens left or right. The standard form for a parabola with vertex \((h,k)\) and opening right/left is \((y - k)^2=4p(x - h)\), where \(p\) is related to latus rectum. The length of latus rectum is \(|4p|\).

Step2: Find \(p\) from latus rectum

Given length of latus rectum is \(\frac{1}{3}\), so \(|4p|=\frac{1}{3}\). Since \(a>0\) (and for opening right, \(p>0\)), \(4p=\frac{1}{3}\), so \(p=\frac{1}{12}\).

Step3: Substitute vertex and \(p\) into equation

Vertex \((h,k)=(4,2)\), \(p=\frac{1}{12}\). Substitute into \((y - k)^2=4p(x - h)\):
\((y - 2)^2=4\times\frac{1}{12}(x - 4)\)
Simplify: \((y - 2)^2=\frac{1}{3}(x - 4)\)

Answer:

\(y^2 = \frac{1}{2}x\)

Problem 22: