
Conic
Sections 


Ellipse

Equation of the ellipse, standard
equation of the ellipse

Major axis, minor axis, and
vertices

The focal parameter, latus rectum

The equation of the ellipse examples 





Equation of the ellipse, standard
equation of the ellipse

If in the direction of axes we introduce a coordinate system so that the center of the ellipse coincides with the 
origin, then coordinates of foci are 
F_{1}(c, 0) and
F_{2}(
c, 0). 
For every
point P(x,
y) of the ellipse, according to 
definition
r_{1} + r_{2} =
2a, it follows
that 

after
squaring 




and
reducing 


Repeated squaring and grouping gives 
(a^{2}

c^{2}) · x^{2}
+ a^{2}y^{2} = a^{2}
· (a^{2} 
c^{2}), 
and
since 
a^{2}

c^{2} = b^{2} 


follows 
b^{2}x^{2}
+ a^{2}y^{2} = a^{2}b^{2} 
equation
of the ellipse, 




and after division by
a^{2}b^{2}, 

standard equation of the ellipse. 


It follows from the equation that an ellipse is defined by values of
a
and b, or as they are associated through
the relation a^{2}

c^{2} = b^{2},
we can say that it is defined by any pair of these three quantities. 
Intersections of an ellipse and the coordinate axes we determine from equation 

by
putting, 

y
= 0 =>
x
= + a,
so obtained are vertices at the ends of the major axis A_{1}(a,
0) and
A_{2}(a,
0), and 
x
= 0 =>
y = + b,
obtained are covertices, the endpoints of the minor axis B_{1}(0,
b) and
B_{2}(0,
b). 
The line segments
A_{1}A_{2
}= 2a and
B_{1}B_{2
} = 2b are the
major and
minor axes while
a and b
are the
semimajor and
semiminor axes respectively. So the arc of the radius
a
centered at B_{1}and
B_{2 }intersects the major axis at
the foci F_{1}and
F_{2}. 
The focal parameter, called latus rectum and denoted
2p, is the chord perpendicular to the major axis
passing through any of the foci, as shows the above figure. The length of which equals the absolute value of ordinates of the points of the ellipse whose abscissas x
= c
or x
= c
that is 

so
the length of the latus rectum 

the
length of the semilatus rectum of the ellipse. 


The
equation of the ellipse examples 
Example:
Given is equation of the ellipse
9x^{2}
+ 25y^{2} = 225, find the lengths of semimajor and
semiminor axes, coordinates of the foci, the eccentricity and the length of the
semilatus rectum. 
Solution: From the standard equation 

we can find the semiaxes
lengths dividing the given 

equation by
225, 


coordinates of the
foci F_{1}(c, 0) and
F_{2}(
c, 0),
since




Example:
From given quantities of an ellipse determine remaining unknown quantities and write equation
of the ellipse, 

Solution:
a) Using 


therefore,
the semiminor axis 


the
linear eccentricity 


the
semi latus rectum 

and
the equation of the ellipse 





the eccentricity 

and
the equation of the ellipse 




the
semi latus rectum 

and
the equation of the ellipse 



d) unknown quantities expressed through given values, 


Example:
Find the equation of
the ellipse whose focus is F_{2}(6,
0) and which passes through the point
A(5Ö3,
4). 
Solution:
Coordinates of the point
A(5Ö3,
4) must satisfy equation of the ellipse, therefore 


thus,
the equation of the ellipse 




Example:
Write equation of the ellipse passing through points
A(4,
2) and B(8,
1). 
Solution:
Given points must satisfy equation of the ellipse, so 

Therefore,
the equation of the ellipse 

or
x^{2}
+ 16y^{2} = 80. 


Example:
In the ellipse 4x^{2}
+ 9y^{2} = 144 inscribed is a rectangle whose vertices lie on the ellipse and
whose sides are parallel with ellipse axes. Longer side, which is parallel to the major axis, relates to the shorter
side as 3 :
2. Find the area of the rectangle. 
Solution: It follows from the given condition that the coordinates
of vertices of the rectangle must satisfy the 
same ratio, i.e.,
x :
y
= 3 :
2
=>
x =
3y/2.

To determine points of the ellipse of which coordinates are in this ratio, put these variable
coordinates into equation of the ellipse, 
P(3y/2,
y) =>
4x^{2}
+ 9y^{2} = 144 
4(3y/2)^{2}
+ 9y^{2} = 144 =>
18y^{2} = 144,

y_{1,2} =
±Ö144/18 =
±2Ö2,
x =
3y/2
=>
x_{1,2} =
±3Ö2.

Therefore, the vertices of the rectangle,




A(3Ö2,
2Ö2),
B(3Ö2,
2Ö2),
C(3Ö2,
2Ö2)
and D(3Ö2,
2Ö2). 
The
area of the rectangle A =
4 ·
x
·
y =
4 ·
(3Ö2
)
· (2Ö2
)
=
48 square units. 








Intermediate
algebra contents 



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