Two Foci and Two Directrices of the Hyperbola

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

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

x2a2y2b2 = 1

⇒ b2x2 - a2y2 = a2b2

Now form the above diagram we get,

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

Two Foci and Two Directrices of the Hyperbola

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(e21) x2 - a2y2 = a2 ∙  a2(e21), [Since, b2 = a2(e21)]

x2(e21) - y2 = a2(e21) = a2e2 - a2

x2e2 - x2 - y2 = a2e2 - a2

x2e2 + a2 + 2 xe a = x2 + a2e2 + 2 x ae x  + y2

(ex + a)2 = (x + ae)2 + y2


(x + ae)2 + y2 = (ex + a)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 hyperbola 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 hyperbola 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, a hyperbola has two foci and two directrices.

The Hyperbola






11 and 12 Grade Math 

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