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2
2
2
if both sides are divided by ( − ), will get ;
2 − 2 = 1
2
2
2 ( − )
In the hyperbola there is a provision that
2
2
2
( – ) = , and if the point
( 0, 0) is executed, it will get equation
2 2
(1) becomes − = 1. which is the
2 2
equation of the hyperbola with the x-axis
(real axis) and y-axis (imaginary axis) as
the axis of symmetry. Hyperbole cuts real
axes at two points, called the vertices of the
hyperbola, namely (− , 0) and ( , 0). The
distance between the two peaks is 2a. The
2 2
foci are 1(− , 0) and 2( , 0). The asymptotes are − = 0 or = and =
2 2

− where ( – ) = .
2
2
2

The hyperbola on the right is centered at M(α, ). Create a new axis X' and Y' that
intersect at M. The relationship between the new axis and the old axis is:
′ = − and = − . If you use the new axis, you will get The equation for
′
′ 2 ′ 2
the hyperbola is − = 1. which if you use the old axis, you will get the equation
2 2

the hyperbola is ( − ) 2 − ( − ) 2 = 1. It was discussed earlier that the locus of the points
2 2
whose distance ratio is with respect to a certain point and a certain line is a fixed value,
ie e = 1, then Kindergarten is in the form of . If 0 < e < 1' then the TK is an ellipse.

ACTIVITY 4

1. One of the vertices of the hyperbola ( −5) 2 − ( −2) 2 = 1 is ….?
16 9

Solutrion :

Given =. . . , =. . ., and the center is at (…. , …..)
Since the coefficients in the hyperbola equation are positive, this hyperbola
2
includes a horizontal hyperbola and has a vertex at . For , we get the vertices of the
hyperbola at (5 ± , 2). And we get the peak of the hyperbola at (… , … ) and
(… , … ).

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