Precalculus question

Question

From the given equation of a parabola x²- 2x - 4y + 32 = 0 , Find the following

1. Coordinates of the vertex
2. Coordinates of the focus
3. Equation of the Latus rectum



I find it hard to understand when there's no solution so please include that thank you

Answer 1

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Answer 2

1. Let put it in standard form: x² - 2x = 4y - 32; x² - 2x + 1 = 4y - 32 + 1; (x - 1)² = 4(y - 31/4

So, vertex of parabola at point (1, 31/4)

2.So focus F(1, 31/4 + 1) = F(1, 35/4)

3.Equation of latus rectum y = 35/4

Answer 3

To do this you need to put this in conic form which is 4p(y-k) = (x-h)^2 when p is focal distance and vertex is at (h,k) so x^2 -2x -4y +32=0 x^2-2x = 4y-32 need to complete the square on the left side. Read up on this if you dont know how to do.

(x-1)^2 -1= 4y-32

(x-1)^2 = 4y-31

(x-1)^2 =4(y-7.75) So vertex is at ( 1, 7.75) 4p=4 so p=1 (focal distance) which would put the focus at (1, 8.75) Drawing a picture of the parabola should help you see this

The latus rectum then is y= 8.75

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