Respuesta :
distance from focus to vertex=distance from vertex to directix
from (6,2) to y=1, the distance is 1
the focus is 1/2 higher than the vertex
vertex=(6,1.5)
for
(x-h)^2=4p(y-k)
vertex=(h,k)
p=distance from vertex to directix
vertex=(6,1.5)
ditance=0.5
(x-6)^2=4(0.5)(y-1.5)
(x-6)^2=2(y-1.5) is the equation
or
y=0.5x^2-6x+19.5 or
y=0.5(x-6)^2+1.5
from (6,2) to y=1, the distance is 1
the focus is 1/2 higher than the vertex
vertex=(6,1.5)
for
(x-h)^2=4p(y-k)
vertex=(h,k)
p=distance from vertex to directix
vertex=(6,1.5)
ditance=0.5
(x-6)^2=4(0.5)(y-1.5)
(x-6)^2=2(y-1.5) is the equation
or
y=0.5x^2-6x+19.5 or
y=0.5(x-6)^2+1.5
Answer:
[tex](y-\frac{3}{2})^2=2(x-6)^2[/tex] equation of parabola
Step-by-step explanation:
Focus is (h,k+p)=(6,2)
And directrix y=k-p=1
When comparing the values we get
h=6 and
k+p=2 (a)
k-p=1 (b)
For solving (a) and (b) we substitute k=2-p in (b) we get:
[tex]2-p-p=1[/tex]
[tex]2-2p=1[/tex]
[tex]p=\frac{1}{2}[/tex]
Hence, put [tex]p=\frac{1}{2}[/tex] in k=2-p we get:
[tex]k=2-\frac{1}{2}[/tex]
[tex]k=\frac{3}{2}[/tex]
Now, we have general equation as:
[tex](y-k)^2=4p(x-h)^2[/tex]
On substituting the values in general equation we get:
[tex](y-\frac{3}{2})^2=4\frac{1}{2}(x-6)^2[/tex]
[tex](y-\frac{3}{2})^2=2(x-6)^2[/tex]