# Directrix of Hyperbola – Equation and Formula

Here you will learn formula to find the equation of directrix of hyperbola with examples.

Let’s begin –

## Directrix of Hyperbola Equation

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

The equation of directrix is x = $$a\over e$$ and x = $$-a\over e$$

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

The equation of directrix is y = $$b\over e$$ and y = $$-b\over e$$

Example : For the given ellipses, find the equation of directrix.

(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 equation of directrix is x = $$\pm {a\over e}$$

$$\implies$$  x = $$\pm {9\over 5}$$

(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 equation of directrix is x = $$\pm {a\over e}$$ + h

$$\implies$$  x = $$\pm {16\over 5}$$ + 1

$$\implies$$ x= $$21\over 5$$ and x = $$-11\over 5$$

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 equation of directrix is x = $$\pm {a\over e}$$ + h

(ii) For conjugate hyperbola,

The equation of directrix is y = $$\pm {b\over e}$$ + k