Question

What are the vertices of the hyperbola given by the equation ((x-3)^2)/(4) - ((y+7)^2)/(1)=1 ?

Answer 1

the center you can read off from the equation, it is \((3,-7)\) and we know it looks like this

Answer 2

ok crappy picture, but you get the idea. i t looks like that because the denominator of the \(x\) term is larger then the denominator of the \(y\) part, and that also gives you the vertex

Answer 3

since \(a^2=4\) we know \(a=2\) so you need to go two units right and left of the center
namely \((1,-7)\) and \((5,-7)\)

Answer 4

thank you so much, I completely understand now. would you mind helping me on a different type of question and working me through it? I've been having trouble with it.

What are the vertices of the hyperbola given by the equation 16x2 – y2 + 96x + 10y + 103 = 0?

Answer 5

yeah we have to write it like the other one

Answer 6

\[16x^2+96x-y^2+10y=-103\] is a start then we have to "complete the square"

Answer 7

\[16(x^2+6x)-(y^2-10y)=-103\]
\[16(x+3)^2-(y-5)^2=-103-5^2+16\times 3^2\] is how you complete the square

Answer 8

you get
\[16(x+3)^2-(y-5)^2=16\] divide by \(16\) and get
\[(x+3)^2-\frac{(y-5)^2}{16}=1\] and now we proceed as before

Answer 9

center is \((-3,5)\) one unit to the left gives \((-4,5)\) and one unit to the right gives \((-2,5)\)

Answer 10

btw i said something incorrect earlier, you know it looks the way it does because the x part is first and the y part is second, not because of the denominators
i was thinking about an ellipse

Answer 11

\[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\] always looks like this

Answer 12

do they really always appear like that?