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
11 and 12 Grade Math
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