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Year 11 2u locus of parabola question (1 Viewer)
- Thread starter kpad5991
- Start date
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')
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')