Find foci, vertices and asymptotes of the hyperbola.

[StudentofSci]

Homework Statement

Find the asymptotes, vertices, and foci of the hyperbola. 4x^2-y^2-24x-4y+28=0

Homework Equations

(x-h)^2/a^2-(y-k)^2/b^2=1, asymptotes= k± (b/a)(x-h), vertices ± a from center, foci ± c from center, c^2=a^2+b^2, center= (h,k).

The Attempt at a Solution

I begin by rewriting the equation of 4x^2-y^2-24x-4y+28=0.
as: 4x^2-24x-y^2-4y=-28
Then I complete the square for each segment (x/y) which ends up as 4(x-3)^2-9 and -(y+2)^2-4.
Rewrite this into the orginal equation as 4(x-3)^2-(y+2)^2=-28+36-4
4(x-3)^2-(y+2)^2=4
I then rewrite this in the form of (x-h)^2/a^2-(y-k)^2/b^2=1
thus (x-3)^2/1-(y+2)^2/4=1
I have now found that a^2=1 and b^2=4 thus, a=1, and b=2
Now I solve for the questions asked
I have the center at (h,k) being (3, -2)
C^2=a^2+b^2, =1+4=5, C=√5
foci are equal to (3±√ 5, -2)
a=1 vertices (4,-2) & (2,-2)
lastly the asymptotes for the form k ± b/a (x-h)
would be in the form of -2 ± 2/1 (x-3), -2 ± 2(x-3)

That is all my work/solutions for this problem.
Any help with what I may have done wrong and how to learn how to correct is, or even just saying "correct" is appreciated thank you.

 

[mighty2000]

Hello! Your work looks good overall. I just have a few suggestions for improvement.

Firstly, when completing the square for each segment, it should be (x-3)^2-9 and -(y+2)^2+4, as you are adding and subtracting these values to both sides of the equation.

Secondly, when rewriting it into the form of (x-h)^2/a^2-(y-k)^2/b^2=1, it should be 4(x-3)^2/9-(y+2)^2/4=1. You made a small error here by not including the denominator 9 for the x term.

Lastly, for the asymptotes, the correct form is y = k ± b/a (x-h), so it should be y = -2 ± 2/1 (x-3) or y = -2 ± 2(x-3).

Other than that, your solutions for the center, foci, and vertices are all correct. Keep up the good work!