We will discuss the hyperbola of the ellipse along with the examples.

The center of a tapered section is a point that bisects any chord that passes through it.

Definition of the center of the hyperbola:

The midpoint of the line segment that the corner points of. connects Hyperbola is called its center.

Assume the equation of the Let the hyperbola be ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1 then observe from the figure above we that C is the center of the line segment AA ‘, where A and A’ are the two corner points. In case of hyperbole ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1, every chord is added to C. halved (0, 0)

Hence, C is the center of the Hyperbola and its coordinates are (0, 0).

Solved examples to find the center of a hyperbola:

**1.** Find the coordinates of the center of the hyperbole 3x (^ {2} ) – 2y (^ {2} ) – 6 = 0.

**Solution:**

The given equation of hyperbole is 3x (^ {2} ) – 2y (^ {2} ) – 6 = 0.

Now form the above equation that we get,

3x (^ {2} ) – 2y (^ {2} ) – 6 = 0

⇒ 3x (^ {2} ) – 2y (^ {2} ) = 6

Now if we divide both sides by 6 we get

( frac {x ^ {2}} {2} ) – ( frac {y ^ {2}} {3} ) = 1 ………… .. (me)

This equation has the form ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1 (a (^ {2} )> b (^ {2} )).

Clearly the center of the hyperbole (1) is at the origin.

Hence the coordinates of the center of the hyperbole 3x (^ {2} ) – 2y (^ {2} ) – 6 = 0 is (0, 0)

**2.**

Find the coordinates of the center hyperbole 5x (^ {2} ) – 9y (^ {2} ) – 10x + 90y + 185 = 0.

**Solution: **

The given equation of hyperbole is 5x (^ {2} ) – 9y (^ {2} ) – 10x – 90y – 265 = 0.

Now form the above equation that we get,

5x (^ {2} ) – 9y (^ {2} ) – 10x – 90y – 265 = 0

⇒ 5x (^ {2} ) – 10x + 5 – 9y (^ {2} ) – 90y – 225 – 265 – 5 + 225 = 0

⇒ 5 (x (^ {2} ) – 2x + 1) – 9 (y (^ {2} ) + 10y + 25) = 45

⇒ ( frac {(x – 1) ^ {2}} {9} ) – ( frac {(y + 5) ^ {2}} {5} ) = 1

We know the equation of hyperbole with the center at (α, β) and the major and minor axes parallel to the x and y axes, ( frac {(x – α) ^ {2}} {a ^ {2}} ) – ( frac {(y – β) ^ {2}} {b ^ {2}} ) = 1.

Now compare the equation ( frac {(x – 1) ^ {2}} {9} ) – ( frac {(y + 5) ^ {2}} {5} ) = 1 with equation ( frac {(x – α) ^ {2}} {a ^ {2}} ) – ( frac {(y – β) ^ {2}} {b ^ {2}} ) = 1 we get

α = 1, β = – 5, a (^ {2} ) = 9 ⇒ a = 3 and b (^ {2} ) = 5 b = √5.

Hence the coordinates of its center are (α, β), ie (1, – 5).

**● ****The ****hyperbole**

**11th and 12th grade math**

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