Answer:
[tex](x-5)^2=4(y+2)[/tex]
Please note that alternative forms of this equation (vertex and standard) are in the explanation.
Step-by-step explanation:
Standard form of a parabola with a vertical axis of symmetry:
[tex]\boxed{(x-h)^2=4p(y-k) \quad \textsf{where}\:p\neq 0}[/tex]
- Vertex = (h, k)
- Focus = (h, k+p)
Given:
- Vertex = (5, -2)
- Focus = (5, -1)
Therefore:
Calculate p:
[tex]\implies -2+p=-1[/tex]
[tex]\implies p=-1+2[/tex]
[tex]\implies p=1[/tex]
Substitute the values of h, k and p into the formula to create an equation of the parabola with the given parameters:
[tex]\implies (x-5)^2=4(1)(y-(-2))[/tex]
[tex]\implies (x-5)^2=4(y+2)[/tex]
In vertex form:
[tex]\implies y=\dfrac{1}{4}(x-5)^2-2[/tex]
In ax² + bx + c form:
[tex]\implies y=\dfrac{1}{4}x^2-\dfrac{5}{2}x+\dfrac{17}{4}[/tex]
