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Natasha2012 [34]
3 years ago
15

Identify all the hyperbolas which open horizontally

Mathematics
2 answers:
dimaraw [331]3 years ago
6 0

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


Mekhanik [1.2K]3 years ago
6 0

Answer:

\frac{(x-2)^2}{3^2}-\frac{(2y-10)^2}{8^2}=1

and

\frac{(x-1)^2}{6^2}-\frac{(2y+6)^2}{5^2}=1

Step-by-step explanation:

There are 2 types of hyperbolas:

Horizontal:

\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1

Vertical:

\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1

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

\frac{(x-2)^2}{3^2}-\frac{(2y-10)^2}{8^2}=1

and

\frac{(x-1)^2}{6^2}-\frac{(2y+6)^2}{5^2}=1

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We are going to make simultaneous equations.
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Now I will use the distributive law on the 3 and what's in the parentheses:
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