# Vertices and Center of Hyperbola Coordinates

Here you will learn how to find the coordinates of the vertices and center of hyperbola formula with examples.

Let’s begin –

## Vertices and Center of Hyperbola Coordinates

#### (i) For the hyperbola $$x^2\over a^2$$ – $$y^2\over b^2$$ = 1

The coordinates of vertices are (a, 0) and (-a, 0).

And the coordinates of center is (0, 0)

#### (ii) For the conjugate hyperbola -$$x^2\over a^2$$ + $$y^2\over b^2$$ = 1

The coordinates of vertices are (0, b) and (0, -b).

And the coordinates of center is (0, 0)

Example : For the given hyperbola, find the coordinates of vertices and center

(i)  $$16x^2 – 9y^2$$ = 144

(ii)  $$9x^2 – 16y^2 – 18x + 32y – 151$$ = 0

Solution :

(i)  We have,

$$16x^2 – 9y^2$$ = 144 $$\implies$$ $$x^2\over 9$$ – $$y^2\over 16$$ = 1,

This is of the form $$x^2\over a^2$$ – $$y^2\over b^2$$ = 1

where $$a^2$$ = 9 and $$b^2$$ = 16 i.e. a = 3 and b = 4

Center of hyperbola is (0, 0)

And Vertices of hyperbola is (a, 0) and (-a, 0).

$$\implies$$  (3, 0) and (-3, 0)

(ii) We have,

$$9x^2 – 16y^2 – 18x + 32y – 151$$ = 0

$$\implies$$ $$9(x^2 – 2x)$$ – $$16(y^2 – 2y)$$ = 151

$$\implies$$  $$9(x^2 – 2x + 1)$$ – $$16(y^2 – 2y + 1)$$ = 144

$$\implies$$  $$9(x – 1)^2$$ – $$16(y – 1)^2$$ = 144

$$\implies$$  $$(x – 1)^2\over 16$$ – $$(y – 1)^2\over 9$$ = 1

Here, a = 4 and b = 3

Coordinates of Center of the hyperbola is (h, k) i.e. (1, 1)

And Coordinates of Vertices are (a + h, k) and (-a + h, k)

$$\implies$$ (4 + 1, 1) and (-4 + 1, 1) = (5, 1) and (-3, 1)

Note : For the hyperbola $$(x – h)^2\over a^2$$ – $$(y – k)^2\over b^2$$ = 1 with center (h. k),

(i) For normal hyperbola,

The coordinates of vertices are (a + h, k) and (-a + h, k).

And Coordinates of Center is (h, k).

(ii) For conjugate hyperbola,

The coordinates of vertices are (h, b + k) and (h, -b + k).

And Coordinates of Center is (h, k).