Math Question (1 Viewer)

[atar90plus]

Hello Guys

Could you guys please help me with this question.

Consider the parabola (y-1)^2 =8(x+2). This question is asking to find the coordinate of the vertex and focus. Although I do not want you to solve this problem. Instead could you guys please

1. Provide me with suggestions on what some of the errors a student may make while doing this question
2. Provide me with the key concepts of this question

Thanks in advance

 

[D94]

Novice mistakes, but I'd say:
- writing the vertex incorrectly (ie. probably taking the point -h,-k instead of h,k)
- likewise with the focus
- getting confused with y² or x²
- drawing a diagram from the vertex on the wrong axis
- getting confused with 4a so the diagram might be upside down


The question is in the form of y² = 4a(x)

We can extend this to a given vertex, ie. (y-k)² = 4a(x-h)

We calculate the focal length through the 4a part, be careful if it's a negative parabola.

Then if we know the vertex, then the focus is within the concave part, so you subtract/add a certain x or y distance (depending on shape) to the vertex. (e.g. h+a, k)

 

[atar90plus]

Thanks

 

[Carrotsticks]

I'm quite prone to silly mistakes and these questions were such little buggers.

So the FIRST thing I did was recognise which type of parabola this was (normal or sideways? positive or negative?), then I drew a GENERAL sketch of what it looked like, regardless of whether the location of the vertex is correct or not. As long as got the shape right.

Then I found the features etc and found the focus and directrix to match that of my diagram (is it on the left/right, do I go up/down?)