We will discuss the definition of ellipse and how to find the equation of the ellipse whose focus, directrix and eccentricity are given.
An ellipse is the locus of a point P moves on this plane in such a way that its distance from the fixed point S always bears a constant ratio to its perpendicular distance from the fixed line L and if this ratio is less than unity.
An ellipse is the locus of a point in a plane which moves in the plane in such a way that the ratio of its distance from a fixed point (called focus) in the same plane to its distance from a fixed straight line (called directrix) is always constant which is always less than unity.
The constant ratio usually denoted by e (0 < e < 1) and is known as the eccentricity of the ellipse.
If S is the focus, ZZ' is the directrix and P is any point on the
ellipse, then by definition
\(\frac{SP}{PM}\) = e
⇒ SP = e ∙ PM
The fixed point S is called a Focus and the fixed straight line L the corresponding Directrix and the constant ratio is called the Eccentricity of the ellipse.
Solved example to find the equation of the ellipse whose focus, directrix and eccentricity are given:
Determine the equation of the ellipse whose focus is at (-1, 0), directrix is 4x + 3y + 1 = 0 and eccentricity is equal to \(\frac{1}{√5}\).
Solution:
Let S (-1, 0) be the focus and ZZ' be the directrix. Let P (x, y) be any point on the ellipse and PM be perpendicular from P on the directrix. Then by definition
SP = e.PM where e = \(\frac{1}{√5}\).
⇒ SP\(^{2}\) = e\(^{2}\) PM\(^{2}\)
⇒ (x + 1)\(^{2}\) + (y - 0)\(^{2}\) = \((\frac{1}{\sqrt{5}})^{2}[\frac{4x + 3y + 1}{\sqrt{4^{2} + 3^{2}}}]\)
⇒ (x + 1)\(^{2}\) + y\(^{2}\) = \(\frac{1}{25}\)\(\frac{4x + 3y + 1}{5}\)
⇒ x\(^{2}\) + 2x + 1 + y\(^{2}\) = \(\frac{4x + 3y + 1}{125}\)
⇒ 125x\(^{2}\) + 125y\(^{2}\) + 250x + 125 = 0, which is the required equation of the ellipse.
● The Ellipse
11 and 12 Grade Math
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