What is the parabola in standard form with a focus of (1/2, 0) and vertex of (0, 0)?

Question

Please help with this question. I need this to study for my math test tomorrow. Thank you for all your help.

Answer 1

We have:  Vertex V at (0,0)  and  Focus F at (1/2, 0)

 

Noting that the focus (1/2, 0), when graphed, is to the right of

the vertex (0,0) this is a parabola that opens to the right instead

of either up or down.

This means the square term in the equation will be the y term

instead of the x term.  The standard form will be in the form of:

x = ay²+by+c

 

That being said, the form we must deal with to work with the

given information is the vertex form of:  (y-k)² = 4p(x-h)

 

(h,k) is the vertex (0,0) so h=0, k = 0

 

p is the distance from the vertex to the focus:

p is positive if the focus is to the right of the vertex and

negative if the focus is to the left of the vertex.

In this case p will be given as a positive value.

Given the vertex (h,k) is at (0,0)

We can see that the distance from (0,0) to (1/2, 0) is 1/2 ...

but, for other problems you need to be aware of the distance

formula:  To find the distance d from (x₁, y₁) to (x₂, y₂):

 

d = √[(x₂-x₁)²+(y₂-y₁)²] = √[(1/2 -0)²+(0-0)²] = √(1/4) = 1/2

 

We have (h,k) = (0,0), p = 1/2 and formula  (y-k)² = 4p(x-h)

 

(y-0)² = 4(1/2)(x-0)

y²= 2x  is the equation of the parabola.

 

Putting this into standard form:

 

x = y²/2    or   x = (1/2)y²

 

Hope this helps.  Let me know if there are questions.  Good luck.