Two Foci and Two Directrices of the Ellipse

We will learn how to find the two foci and two directrices of the ellipse.

Let P (x, y) be a point on the ellipse.

x2a2 + y2b2 = 1

⇒ b2x2 + a2y2 = a2b2

Now form the above diagram we get,

CA = CA' = a and e is the eccentricity of the ellipse and the point S and the line ZK are the focus and directrix respectively.

Now let S' and K' be two points on the x-axis on the side of C which is opposite to the side of S such that CS' = ae and CK' = ae.

Further let Z'K' perpendicular CK' and PM' perpendicular Z'K' as shown in the given figure. Now join P and S'. Therefore, we clearly see that PM’ = NK'.

Now from the equation b2x2 + a2y2 = a2b2, we get,

⇒ a2(1 - e2) x2 + a2y2 = a2 . a2(1 - e2), [Since, b2 = a2(1 - e2)]

⇒ x2(1 -  e2) + y2 = a2(1 - e2) = a2 – a2e2

⇒ x2 + a2e2 + y2 = a2 + x2e2

⇒ x2 + (ae)2 + 2 ∙ x ∙ ae + y2 = a2 + x 2e2 + 2a ∙ xe

⇒ (x + ae)2 + y2 = (a + xe)2

⇒ (x + ae)2 + (y - 0)2 = e2(x + ae)2

⇒ S'P2 = e2 ∙ PM'2

⇒ S'P = e ∙ PM'

Distance of P from S' = e (distance of P from Z'K')

Hence, we would have obtained the same curve had we started with S' as focus and Z'K' as directrix. This shows that the ellipse has a second focus S' (-ae, 0) and a second directrix x = -ae.

In other words, from the above relation we see that the distance of the moving point P (x, y) from the point S' (- ae, 0) bears a constant ratio e (< 1) to its distance from the line x + ae = 0.

Therefore, we shall have the same ellipse if the point S' (- ae, 0) is taken as the fixed point i.e, focus and x + ae = 0 is taken as the fixed line i.e., directrix.

Hence, an ellipse has two foci and two directrices.

● The Ellipse




11 and 12 Grade Math 

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