Identify all the hyperbolas which open horizontally

Answer:
The hyperbolas which open horizontally are:
(x+2)^2/3^2-(2y-10)^2/8^2=1
(x-1)^2/6^2-(2y+6)^2/5^2=1
Step-by-step explanation:
A hyperbola with equation of the form:
(x-h)^2/a^2-(y-k)^2/b^2)=1 opens horizontally
Then, the hyperbolas which open horizontally are:
(x+2)^2/3^2-(2y-10)^2/8^2=1
(x-1)^2/6^2-(2y+6)^2/5^2=1
Answer:
[tex]\frac{(x-2)^2}{3^2}-\frac{(2y-10)^2}{8^2}=1[/tex]
and
[tex]\frac{(x-1)^2}{6^2}-\frac{(2y+6)^2}{5^2}=1[/tex]
Step-by-step explanation:
There are 2 types of hyperbolas:
Horizontal:
[tex]\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1[/tex]
Vertical:
[tex]\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1[/tex]
If the x term is positive then parabola is horizontal.
If the y term is positive then parabola is vertical.
so, only first two equations are in which x term is positive.
hence, equations are
[tex]\frac{(x-2)^2}{3^2}-\frac{(2y-10)^2}{8^2}=1[/tex]
and
[tex]\frac{(x-1)^2}{6^2}-\frac{(2y+6)^2}{5^2}=1[/tex]