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joja [24]
3 years ago
9

Derive the equation of the parabola with a focus at (−5, 5) and a directrix of y = −1.

Mathematics
2 answers:
marusya05 [52]3 years ago
7 0

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

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,

(x-h)^2=4p(y-k)

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).

(h,k+p)=(-5,5)

On comparing both sides we get,

h=-5

k+p=5      .... (1)

The directrix of y = −1.

k-p=-1       .... (2)

Add equation (1) and (2),

2k=4

k=2

Put this in equation (1),

p=3

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

(x-(-5))^2=4(3)(y-2)

(x+5)^2=12(y-2)

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

Sauron [17]3 years ago
6 0
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.

\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}\\\\
-------------------------------\\\\

\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

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