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 = $-b\over e$

Also Read : Different Types of Ellipse Equations and Graph

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