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
