Year 11 2u locus of parabola question (1 Viewer)

[kpad5991]

Find all possible equations of the parabola with a focus of (6,-3) and the vertex is on the line y= x-4.
(Assuming that the axis is parallel to one of the coordinate axes)

 

[Sp3ctre]

I'm a bit rusty with locus so hopefully I haven't made any mistakes in this.

Draw a graph of y=x-4 and plot the point (6,-3) first, You should see that the parabola's vertices can only possibly be located at the ordinates x=6 or y=-3. Find the x and y coordinates by subbing x=6 and y=-3 separately into y=x-4

When x=6
y = 6-4
y= 2
Thus one possible vertex is at (6,2)

When y=-3
-3 = x - 4
x = 1
Thus the other possible vertex is at (1,-3)

Then just sub those points into the equation of a parabola
(x-6)^2 = -4a(y-2)

(y+3)^2 = 4a(x-1)

(By inspection, the vertical parabola is concave down and sideways parabola is 'concave right')