Geometry question

Question

A parabola can be drawn given a focus of (−9,1)(−9,1) and a directrix of x=−3x=−3. What can be said about the parabola?

Answer 1

The axis of symmetry is perpendicular to the directrix and passes through the focus. So, the axis of symmetry is the horizontal line y = 1.The vertex of the parabola lies on the line y = 1 and is halfway between the focus and directrix. So, the vertex is (-6,1). The focus is inside the arc of the parabola, so the parabola opens to the left.

Equation has the form x = a(y - 1)² - 6, where a = 1/ (4p). Since the focus lies 3 units to the left of the vertex, p = -3.

So, x = (-1/12)(y - 1)² - 6

Answer 2

directrix is x=-3

focus is (-9,1)

vertex is (-6,1) which is half way between the focus and the vertex (-3-9)/2=-12/2=-6

it's a leftward opening parabola

the general equation for the leftward opening parabola in vertex form is: x=a(y-k)^2 +h where (h,k) is the vertex

plug in the vertex. x=a(y-1)^2 -6

plug in another point on the parabola and solve for the coefficient "a" which is <0 for a leftward opening parabola

two other points on the parabola are directly above and below the focus:

equidistant to the directrix and focus= 6, (-9,1+6) and (-9,1-6) =

(-9, 7) and (-9,-5)

-9= a(6^2) -6

-3 = 36a

a =-3/36= -1/12=-1/4p with p=6/2 = 3

x = (-1/12)(y-1)^2 -6

or

12x = -y^2 +2y -1 -72

y^2 -2y +12x = -73

when y=0, x = -73/12 = the x intercept = (-73/12, 0) =(-6 1/12, 0)

there are no y intercepts

when x=0 solve for y

y= 2/2 + or - (1/2)sqr(-296) which are two imaginary numbers

the parabola is always to the left of the y axis and never intersects it

axis of symmetry is y=1