We will learn how to find the two focal points and two directions of the hyperbola.

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

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

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

Now build the above diagram that we get,

CA = CA ‘= a and e is the eccentricity of the The hyperbola and the point S and the line ZK are the focus or the guideline.

Let S ‘and K’ now be two points on the x-axis on the side of C opposite the side of S, so that CS ‘= ae and CK’ = ( frac {a} {e} ).

Furthermore, let Z’K ‘be perpendicular to CK’ and PM ‘perpendicular to Z’K’, as shown in the figure given. Now connect P and S ‘. Hence we clearly see that PM ‘= NK’.

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 **∙**

aEx + j (^ {2} )

⇒ (e.g. + a) (^ {2} ) = (x + ae) (^ {2} ) + Yes (^ {2} )

⇒ (x + ae) (^ {2} ) + Yes (^ {2} ) = (e.g. + 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**∙**

PN ‘

Distance from P to S ‘= e (distance from P to Z’K’)

Therefore we would have got the same curve if we had started with S ‘as focus and Z’K’ as guideline. This shows that the hyperbole has a second focus S ‘(-ae, 0) and a second guideline x = – ( frac {a} {e} ).

In other words, from the above relationship we see that the distance of the moving point P (x, y) from the point S ‘(- ae, 0) has a constant ratio e (> 1) to its distance from the line x + ( frac {a} {e} ) = 0.

Hence we will have the same hyperbole

if the point S ‘(- ae, 0) is taken as a fixed point, ie focus, and x + ( frac {a} {e} ) = 0 as a fixed point, ie directrix.

Hence a hyperbole has two focal points and two directorates.

**● ****The ****hyperbole**

**11th and 12th grade math**

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