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 xaxis 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
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.