Chapter 2.3, Problem 46E

To determine

To describe: Identify the vertex, focus, directrix and axis of symmetry of the parabola.

Answer to Problem 46E

  $\begin{array}{l}\text{Therefore, }\\ \text{Vertex:}(-1,-5);\\ \text{Focus:}\left(\frac{-15}{16},-5\right);\\ \text{Directrix:}x=-1-\frac{1}{16}=\frac{-17}{16}\\ \text{Axis of symmetry:}y=-5\\ \text{Transformation of the graph, }x=4y^2\end{array}$

Explanation of Solution

Given information: The equation of the parabola is,

  $x=4{(y+5)}^2-1$

Formula Used: The general equation of parabola,

  $x=\frac{1}{4p}{(y-k)}^2+h$

Calculation: To identify the vertex, focus, directrix and axis of symmetry for,

  $\begin{array}{l}$ x=4{(y+5)}^2-1$\text{On comparing with general form}\\ $ x=\frac{1}{4p}{(y-k)}^2+h$\text{where Vertex:}(h,k)\text{ and Focus:(}h+p,k)\\ \text{and Directrix:}x=h-p\\ $ x=4{(y+5)}^2-1$\text{Then, Vertex:}(-1,-5);\\ \text{Focus:}\left(\frac{-15}{16},-5\right);\\ \text{Directrix:}x=-1-\frac{1}{16}=\frac{-17}{16}\\ \text{Axis of symmetry:}y=-5\end{array}$

The transformation of the graph of the standard equation with $p=1\text{and vertex}(0,0).$

Therefore, the equation is: $x=4y^2$