Chapter 11, Problem 4RE

a

To determine

To Find: The vertex, focus and directive of parabola

Answer to Problem 4RE

Vertex= (o, o)

Focus= $(\frac{1}{2},0)$

Directive is 2x+1=0

Explanation of Solution

Given information:

  $2x-y^2=0$

Concept Used:

  $y^2=hax$

Given equation is $\begin{array}{l}2x-y^2=0\\ =y^2=2x\end{array}$

By comparing these equations with standard form we can arrive at required values.

Calculation:

Comparing this with standard form

4a=2

a=0.5

Vertex= origin= (0,0)

Co-ordinates of focus= (a,0)= (0.5,0)

Directive is $\begin{array}{l}x=-a\\ x=-0.5=\frac{-1}{2}\\ 2x+1=0\end{array}$

Hence, vertex $=\left(0,0\right)$

Focus $=\left(0.5,0\right)$

Directive $=2x+1=0$

Conclusion:

The vertex obtained is $=\left(0,0\right)$ , Focus is at $=\left(0.5,0\right)$ and directive is $=2x+1=0$

b.

To determine

Sketch the graph:

To draw the graph of given equation $2x-y^2=0$

Answer to Problem 4RE

The graph $2x-y^2=0$ can be obtained by giving different non-negative values of x

Explanation of Solution

Given information: $\begin{array}{l}2x-y^2=0\\ =y^2=2x\end{array}$

Graph

  $\begin{array}{l}x=0,y=0\\ x=2,y=\pm 2\\ x=2,y=\pm 4\end{array}$

The points on graph would be $\left(0,0\right)\left(2,2\right)\left(2,-2\right)\left(8,4\right)\left(8,-4\right)$

  

Interpretation :

Using these points graph is obtained from $\left(0,0\right)\left(2,2\right)\left(2,-2\right)\left(8,4\right)\left(8,-4\right)$