Respuesta :

check the picture below.

so, the left-part of the picture is the focus point and the directrix... .so.. notice, the focus is above the directrix, meaning is a vertical parabola and is opening upwards.

now, looking at the right-part of the picture, bear in mind that, the vertex is "p" units away from the directrix and the focus, so, the vertex is half-way between those fellows, notice in the picture what "p" is, keep in mind that, because the parabola is opening upwards, "p" is a positive unit, thus is 3.

[tex]\bf \textit{parabola vertex form with focus point distance}\\\\ \begin{array}{llll} (y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\ \boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}}) }\\ \end{array} \qquad \begin{array}{llll} vertex\ ({{ h}},{{ k}})\\\\ {{ p}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\\\\ -------------------------------\\\\[/tex]

[tex]\bf \begin{cases} h=-5\\ k=2\\ p=3 \end{cases}\implies [x-(-5)]^2=4(3)(y-2) \\\\\\ (x+5)^2=12(y-2)\implies \cfrac{(x+5)^2}{12}=y-2\implies \cfrac{(x+5)^2}{12}+2=y \\\\\\ \cfrac{1}{12}(x+5)^2+2=y[/tex]
Ver imagen jdoe0001

Answer: The equation of parabola is [tex](x+5)^2=12(y-2)[/tex].

Explanation:

It is given that the focus of the parabola is (−5, 5) and a directrix of y = −1.

The standard form of a parabola is,

[tex](x-h)^2=4p(y-k)[/tex]

Where, (h,k+p) is the focus and y=k-p is the directrix.

It is given that  the focus of the parabola is (−5, 5).

[tex](h,k+p)=(-5,5)[/tex]

On comparing both sides we get,

[tex]h=-5[/tex]

[tex]k+p=5[/tex]      .... (1)

The directrix of y = −1.

[tex]k-p=-1[/tex]       .... (2)

Add equation (1) and (2),

[tex]2k=4[/tex]

[tex]k=2[/tex]

Put this in equation (1),

[tex]p=3[/tex]

Put p=3, k=2 and h=-5 in standard equation of parabola.

[tex](x-(-5))^2=4(3)(y-2)[/tex]

[tex](x+5)^2=12(y-2)[/tex]

Therefore, the equation of parabola is [tex](x+5)^2=12(y-2)[/tex].

Ver imagen DelcieRiveria

Otras preguntas