Foci of Hyperbola Formula and Coordinates

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).

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).