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)
Also Read : Equation of the Hyperbola | Graph of a Hyperbola
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).
