
Conic
Sections 


Hyperbola

Translated hyperbola

Equation of the hyperbola in vertex form

Parametric equations of the hyperbola

Examples of hyperbola 





Translated hyperbola

The equation of a hyperbola translated from standard position so that its center is at
S(x_{0}, y_{0})
is given by

b^{2}(x

x_{0})^{2}

a^{2}(y

y_{0})^{2}
= a^{2}b^{2} 
or




and after expanding and substituting constants
obtained is

Ax^{2}
+ By^{2}
+ Cx + Dy + F
= 0.

An equation of that form represents the hyperbola if

A · B < 0

that is, if coefficients of the square terms have
different signs.





Equation of the hyperbola in vertex form

By translating the hyperbola, centered at
(0, 0), in the
negative direction of the xaxis by
x_{0}
= a, so that new
position of the center S(a, 0)
then its equation is 
b^{2}(x
+ a)^{2}

a^{2}y^{2}
= a^{2}b^{2}.
After squaring and reducing,

b^{2}x^{2}
+ 2ab^{2}x

a^{2}y^{2}
= 0 or 


since 

obtained is 


the equation of the hyperbola in vertex form.





Parametric equation of the hyperbola

In the construction of the hyperbola, shown in
the below figure, circles of radii
a
and b
are intersected by an arbitrary line through the origin at points M
and N. Tangents to the circles at
M
and N
intersect the xaxis at
R
and S.
On the perpendicular through S, to the
xaxis, mark the line segment
SP of length
MR to get the point
P of the hyperbola.
We can prove that P is a point of the hyperbola.

In the right
triangles ONS and
OMR,


by replacing OS
= x
and MR
= SP

and substituting 


by
dividing by b^{2}, 


therefore, P(x, y)
is
the point of the hyperbola.




The coordinates of the point
P(x, y)
can also be expressed by the angle t
common to both mentioned 
triangles,
so that 

is the parametric equation of the hyperbola. 

By substituting these parametrically expressed coordinates into equation of the hyperbola 

that
is, 

known
trigonometric identity. 


Examples
of hyperbola 
Example:
Given is the hyperbola
4x^{2} 
9y^{2} = 36,
determine the semiaxes, equations of the asymptotes,
coordinates of foci, the eccentricity and the semilatus rectum.

Solution:
Put the equation in the standard form to
determine the semiaxes, thus

4x^{2}

9y^{2} = 36  ¸
36


Asymptotes, 


Applying, 


coordinates of foci, F_{1}(Ö13,
0) and
F_{2}(Ö13,
0).




The eccentricity, 

and the
semilatus rectum,




Example:
Write equation of a hyperbola with the focus at
F_{2}(5,
0) and whose asymptotes are, 




Therefore,
the equation of the hyperbola,





Example:
Find the angle subtended by the focal radii
r_{1}
and
r_{2}
at a point A(8, y
> 0) of the hyperbola

9x^{2}

16y^{2} = 144. 
Solution:
We determine the ordinate of the point
A
by plugging its abscissa into equation of the hyperbola,

x
= 8 =>
9x^{2}

16y^{2} = 144

9 · 8^{2}

16y^{2} = 144,

16y^{2} = 432 =>
y^{2} =
27,

y_{1,2} = ±3Ö3
so that A(8,
3Ö3).

The equations of the lines of the radii r_{1}
and
r_{2}, we
write
using the formula of a line through two
points. Since,

r_{1}::
AF_{1} and
F_{1}(c,
0) ,
and 


then 







Therefore, the angle between the focal radii r_{1}
and
r_{2}
at the point A
of the hyperbola, as



Example:
The hyperbola is given by equation
4x^{2}

9y^{2} + 32x + 54y 
53 =
0. 
Find coordinates of the
center, the foci, the eccentricity and the asymptotes of the hyperbola. 
Solution:
The given hyperbola is translated in the direction of the coordinate axes so the values of
translations x_{0} and
y_{0}
we can find by using the method of completing the square
rewriting the equation in 
the standard
form, 


Thus,
4x^{2} + 32x

9y^{2} + 54y 
53 =
0,

4(x^{2} + 8x) 
9(y^{2} 
6y) 
53 =
0

4[(x +
4)^{2} 
16] 
9[(y 
3)^{2} 
9] 
53 =
0

4(x + 4)^{2} 
9(y 
3)^{2} = 36  ¸
36

Therefore,






it
follows that a^{2} =
9, a
= 3,
b^{2} = 4,
b
= 2,
and the center of the hyperbola at S(x_{0}, y_{0})
or S(4,
3). 
Half
the focal distance 

the
eccentricity 


and the foci, F_{1}(x_{0
}
c,
0) so
F_{1}(4_{
}
Ö13,
0) and
F_{2}(x_{0
} + c,
0),
F_{1}(4_{
} + Ö13,
0). 
Equations of the asymptotes of a translated hyperbola 

therefore, the asymptotes of the given
hyperbola, 



Example:
Write the equation of the hyperbola 9x^{2}

25y^{2} = 225
in the vertex form. 
Solution:
Using parallel shifting we should place the center of the hyperbola at
S(a,
0). 
Rewrite
9x^{2}

25y^{2} = 225  ¸
225 


therefore,
a = 5 and
b = 3,
so that S(5,
0). 

Then, the translated
hyperbola with the center at S(5,
0) has the equation










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