150k views
3 votes
ANYONE PLEASE HELP ME WITH THIS QUESTION AND OTHERS IN MY PROFILE I NEED TO GET FINISHED IM WILLING TO TALK WITH HELPER ON ABOUT SOME TERMS !!

ANYONE PLEASE HELP ME WITH THIS QUESTION AND OTHERS IN MY PROFILE I NEED TO GET FINISHED-example-1

1 Answer

7 votes

Answer:

1) vertex = (-2,4)

2) Focus = (-0.5,4)

3) x= -3.5

y= 4

4) y=4

Explanation:

General equation of parabola that is parallel to a-axis and vertex at (h,k) is given as

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

where

vertex of parabola is at (h,k)

focus of parabola is given at (h + p, k)

the directrix of parabola is given as x = h - p.

Now

1)

finding vertex of parabola:

Given equation of parabola

(y-4)^2=6(x+2)

Comparing with the general form, we get

h=-2 ,k=4 and 4p=6

hence vertex = (-2,4)

2)

Finding focus

Comparing with the above standard form we get

k=4, h=-2, p=3/2

Since the given parabola is parallel to x-axis and also p is positive hence it will opens to the right.

As focus is inside the parabola and it is p units to the right of the vertex:

hence

focus of parabola (h + p, k)=(-2+3/2 , 4)

=(-0.5,4)

3)

Comparing with the above standard form we get

k=4, h=-2, p=3/2

Since the given parabola is parallel to x-axis and also p is positive hence it will opens to the right.

As directrix is outside the parabola and it is p units to the left of the vertex:

hence

directrix x=h-p

= -2-3/2

=-7/2

= -3.5

y= 4

4)

Finding Axis of symmetry:

as the vertex is (-2,4) also the given parabola is parallel to x-axis so

the axis of symmetry is a horizontal straight line passing through the vertex at y=4 !

User Ermintar
by
7.3k points