# Directrix of Ellipse – Equation and Formula

Here you will learn what is the formula to find the equation of directrix of ellipse with examples.

Let’s begin –

## Directrix of Ellipse Equation

#### (i) For the ellipse $$x^2\over a^2$$ + $$y^2\over b^2$$ = 1, a > b

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

#### (ii) For the ellipse $$x^2\over a^2$$ + $$y^2\over b^2$$ = 1, a < b

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)  $$16x^2 + 25y^2$$ = 400

(ii)  $$x^2 + 4y^2 – 2x$$ = 0

Solution :

(i)  We have,

$$16x^2 + 25y^2$$ = 400 $$\implies$$ $$x^2\over 25$$ + $$y^2\over 16$$,

where $$a^2$$ = 25 and $$b^2$$ = 16 i.e. a = 5 and b = 4

Clearly a > b,

The eccentricity of ellipse (e) = $$\sqrt{1 – {b^2\over a^2}}$$

e = $$\sqrt{1 – 16/25}$$ = $$3\over 5$$

Now, the equation of directrix is x = $$a\over e$$ and x = $$-a\over e$$

$$\implies$$  x = $$25\over 3$$ and x = $$-25\over 3$$

(ii) We have,

$$x^2 + 4y^2 – 2x$$ = 0

$$\implies$$ $$(x – 1)^2$$ + 4$$(y – 0)^2$$ = 1

$$\implies$$  $$(x – 1)^2\over 1^2$$ + $$(y – 0)^2\over (1/2)^2$$ = 1

Here, a = 1 and b = 1/2

Clearly a > b,

The eccentricity of ellipse (e) = $$\sqrt{1 – {b^2\over a^2}}$$

e = $$\sqrt{1 – 1/4}$$ = $$\sqrt{3}\over 2$$

Since, center of this ellipse is (1, 0)

Therefore, the equation of directrix is x = $$a\over e$$ + h and x = $$-a\over e$$ + h

$$\implies$$  x = $$2\over \sqrt{3}$$ + 1 and y = $$-2\over \sqrt{3}$$ + 1

Note : For the ellipse $$(x – h)^2\over a^2$$ + $$(y – k)^2\over b^2$$ = 1 with center (h. k),

(i) For ellipse a > b,

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

(ii) For ellipse a < b,

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