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

goldenlink7ydw

goldenlink7ydw

Answered question

2023-02-17

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

Answer & Explanation

sarabol2zsr

sarabol2zsr

Beginner2023-02-18Added 6 answers

the equation of a horizontally opening parabola is
x ( y - k ) 2 = 4 a ( 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 )
4 a = - 1 a = - 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 + 3 x = - 1
for y-intercepts let x = 0
( y - 2 ) 2 = 3 ( y - 2 ) = ± 3
y = 2 ± 3
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]}

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