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.

\(\frac{x^{2}}{a^{2}}\) - \(\frac{y^{2}}{b^{2}}\) = 1

⇒ b\(^{2}\)x\(^{2}\) - a\(^{2}\)y\(^{2}\) = a\(^{2}\)b\(^{2}\)

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.

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' = \(\frac{a}{e}\).

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 b\(^{2}\)x\(^{2}\) - a\(^{2}\)y\(^{2}\) = a\(^{2}\)b\(^{2}\), we get,

⇒ a\(^{2}\)(e\(^{2} - 1\)) x\(^{2}\) - a\(^{2}\)y\(^{2}\) = a\(^{2}\) **∙ ** a\(^{2}\)(e\(^{2} - 1\)), [Since, b\(^{2}\) = a\(^{2}\)(e\(^{2} - 1\))]

⇒ x\(^{2}\)(e\(^{2} - 1\)) - y\(^{2}\) = a\(^{2}\)(e\(^{2} - 1\)) = a\(^{2}\)e\(^{2}\) - a\(^{2}\)

⇒ x\(^{2}\)e\(^{2}\) - x\(^{2}\) - y\(^{2}\) = a\(^{2}\)e\(^{2}\) - a\(^{2}\)

⇒ x\(^{2}\)e\(^{2}\) + a\(^{2}\) + 2 **∙**
xe **∙**
a = x\(^{2}\) + a\(^{2}\)e\(^{2}\) + 2 **∙**
x **∙**
ae x + y\(^{2}\)

⇒ (ex + a)\(^{2}\) = (x + ae)\(^{2}\) + y\(^{2}\)

⇒ (x + ae)\(^{2}\) + y\(^{2}\) = (ex + a)\(^{2}\)

⇒ (x + ae)\(^{2}\) - (y - 0)\(^{2}\) = e\(^{2}\)(x + \(\frac{a}{e}\))\(^{2}\)

⇒ S'P\(^{2}\) = e\(^{2}\) **∙**
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 = -\(\frac{a}{e}\).

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 + \(\frac{a}{e}\) = 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 + \(\frac{a}{e}\) = 0 is taken as the fixed line i.e., directrix.

Hence, a hyperbola has two foci and two directrices.

**● ****The ****Hyperbola**

**Definition of Hyperbola****Standard Equation of an Hyperbola****Vertex of the Hyperbola****Centre of the Hyperbola****Transverse and Conjugate Axis of the Hyperbola****Two Foci and Two Directrices of the Hyperbola****Latus Rectum of the Hyperbola****Position of a Point with Respect to the Hyperbola****Conjugate Hyperbola****Rectangular Hyperbola****Parametric Equation of the Hyperbola****Hyperbola Formulae****Problems on Hyperbola**

**11 and 12 Grade Math**__From Two Foci and Two Directrices of the Hyperbola____ to HOME PAGE__

**Didn't find what you were looking for? Or want to know more information
about Math Only Math.
Use this Google Search to find what you need.**

## New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.