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.

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

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

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

11 and 12 Grade Math 

From Two Foci and Two Directrices of the Hyperbola to HOME PAGE

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.

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.

Share this page: What’s this?

Recent Articles

  1. 2nd grade math Worksheets | Free Math Worksheets | By Grade and Topic

    Dec 06, 23 01:23 AM

    2nd Grade Math Worksheet
    2nd grade math worksheets is carefully planned and thoughtfully presented on mathematics for the students.

    Read More

  2. Rupees and Paise | Paise Coins | Rupee Coins | Rupee Notes

    Dec 04, 23 02:14 PM

    Different types of Indian Coins
    Money consists of rupees and paise; we require money to purchase things. 100 paise make one rupee. List of paise and rupees in the shape of coins and notes:

    Read More

  3. Months of the Year | List of 12 Months of the Year |Jan, Feb, Mar, Apr

    Dec 04, 23 01:50 PM

    Months of the Year
    There are 12 months in a year. The months are January, February, march, April, May, June, July, August, September, October, November and December. The year begins with the January month. December is t…

    Read More