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

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