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 = $-b\over e$

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

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