Question

How do you Find the vertex, Focus, and directrix of the Parabola and sketch its graph `(x-3) + (y-2)^2=0`?

Answer 1

`"see explanation"`

Explanation:

`"the equation of a horizontally opening parabola is"`

`•color(white)(x)(y-k)^2=4a(x-h)`

`"where "(h,k)" are the coordinates of the vertex and a is"`
`"the distance from the vertex to the focus and directrix"`

`"If "a>0" opens to the right "`

`"If "a<0" opens to the left"`

`"rearrange the given equation into this form"`

`(y-2)^2=-(x-3)`

`"with vertex "=(3,2)`

`4a=-1rArra=-1/4" so opens to the left"`

`"Focus "=(a+h,k)=(-1/4+3,2)=(11/4,2)`

`"directrix is "x=-a+h=1/4+3=13/4`

`"for x-intercept let y = 0"`

`4=-x+3rArrx=-1`

`"for y-intercepts let x = 0"`

`(y-2)^2=3rArr(y-2)=+-sqrt3`

`y=2+-sqrt3`

`y~~3.73,y~~0.27`

`"Plot the above coordinates for vertex, focus and "`
`"intercepts and draw a smooth curve through them"`
graph{(y-2)^2=-(x-3) [-10, 10, -5, 5]}