Here you will learn how to find the coordinates of the foci of hyperbola with examples.
Let’s begin –
Foci of Hyperbola Coordinates
(i) For the hyperbola \(x^2\over a^2\) – \(y^2\over b^2\) = 1
The coordinates of foci are (ae, 0) and (-ae, 0).
(ii) For the conjugate hyperbola -\(x^2\over a^2\) + \(y^2\over b^2\) = 1
The coordinates of foci are (0, be) and (0, -be).
Also Read : Equation of the Hyperbola | Graph of a Hyperbola
Example : For the given hyperbola, find the coordinates of foci
(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
Eccentricity (e) = \(\sqrt{1 + {b^2\over a^2}}\)
\(\implies\) e = \(\sqrt{1 + 16/9}\) = \(5\over 3\)
Therefore, the coordinates of foci are (ae, 0) and (-ae, 0).
\(\implies\) (5, 0) and (-5, 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
Eccentricity (e) = \(\sqrt{1 + {b^2\over a^2}}\)
\(\implies\) e = \(\sqrt{1 + 9/16}\) = \(5\over 4\)
Here, the center is (h, k) i.e. (1, 1)
Therefore, the coordinates of foci are (ae + h, k) and (-ae + h, k).
\(\implies\) (6, 1) and (-4, 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 foci are (ae + h, k) and (-ae + h, k).
(ii) For conjugate hyperbola,
The coordinates of foci are (h, be + k) and (h, -be + k).
