arrebolyt

Answered

2022-01-28

What is the equation in standard form of the parabola with a focus at (14,5) and a directrix of y=-15?

Answer & Explanation

Madelyn Townsend

Expert

2022-01-29Added 13 answers

Explanation:

Focus is at (14,5) and directrix is y=-15. Vertex is at midway

between focus and directrix. Therefore vertex is at

(14,$\frac{5-15}{2}$ ) or (14, -5). The vertex form of equation of

parabola is$y=a{(x-h)}^{2}+k$ ; (h.k); being vertex. Here

h=14 and k=-5 So the equation of parabola is

$y=a{(x-14)}^{2}-5$ . Distance of vertex from directrix is

d=15-5=10, we know$d=\frac{1}{4\left|a\right|}\therefore \left|a\right|=\frac{1}{4d}$ or

$\left|a\right|=\frac{1}{4\cdot 10}=\frac{1}{40}$ . Here the directrix is below

the vertex , so parabola opens upward and a is positive.

$\therefore a=\frac{1}{40}$ Hence the equation of parabola is

$y=\frac{1}{40}{(x-14)}^{2}-5$

Focus is at (14,5) and directrix is y=-15. Vertex is at midway

between focus and directrix. Therefore vertex is at

(14,

parabola is

h=14 and k=-5 So the equation of parabola is

d=15-5=10, we know

the vertex , so parabola opens upward and a is positive.

mihady54

Expert

2022-01-30Added 13 answers

Explanation:

the standard form of a parabola in translated form is.

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

where (h,k) are the coordinates of the vertex

and p is the distance from the vertex to the focus

since the directrix is below the focus then the curve

opens upwards

coordinates of vertex$=(14,\frac{5-15}{2})=(14,-5)$

and p=5-(-5)=10

$\Rightarrow {(x-14)}^{2}=40(y+5)\leftarrow$ equation of parabola

the standard form of a parabola in translated form is.

where (h,k) are the coordinates of the vertex

and p is the distance from the vertex to the focus

since the directrix is below the focus then the curve

opens upwards

coordinates of vertex

and p=5-(-5)=10

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