Conic section- ellipse STEM TEACH

1. CONIC SECTION

2. ELLIPSE

3. Given the equation of the ellipse
25𝑥2 + 4𝑦2 + 100𝑥 + 8𝑦 + 4 = 0 find the following:
a. coordinates of the center
b. vertices
c. foci
d. length of minor axis.
e. length of major axis
g. graph the equation.

4. SOLUTION
Since the second degree is raised in x, we expect that the
standard form follows
(𝑥−ℎ)2
𝑎2 +
(𝑦−𝑘)2
𝑏2 = 1

5. 25𝑥2
+ 4𝑦2
+ 100𝑥 + 8𝑦 + 4 = 0
First step: Isolate the constant and group x’s and y’s
25𝑥2
+ 100𝑥 + 4𝑦2
+ 8𝑦 = −4

6. 25𝑥2
+ 100𝑥 + 4𝑦2
+ 8𝑦 = −4
Second step: Factor out 25 and 4
25(𝑥2
+4𝑥) + 4(𝑦2
+2𝑦) = −4

7. 25(𝑥2
+4𝑥) + 4(𝑦2
+2𝑦) = −4
Third step: Undergo completing the square process by adding a
constant 4 and 1 and multiply this to 25 and 4 to add on the right
hand side then add the result.
25(𝑥2
+4𝑥 + _) + 4(𝑦2
+2𝑦 + _) = −4

8. 25(𝑥2
+4𝑥) + 4(𝑦2
+2𝑦) = −4
Third step: Undergo completing the square process by adding a
constant 4 and 1 and multiply this to 25 and 4 to add on the right
hand side then add the result.
25(𝑥2+4𝑥 + _) + 4(𝑦2+2𝑦 + _) = −4
25(𝑥2+4𝑥 + 𝟒) + 4(𝑦2+2𝑦 + 𝟏) = −4 + 𝟏𝟎𝟎 + 𝟒

9. 25(𝑥2+4𝑥 + 𝟒) + 4(𝑦2+2𝑦 + 𝟏) = −4 + 𝟏𝟎𝟎 + 𝟒
Fourth step: Divide both side by 100 making the right side equal
to 1. Simplify the result
25(𝑥+2)2+4(𝑦+1)2
100
=
100
100

10. 25(𝑥+2)2+4(𝑦+1)2
100
=
100
100
(𝑥 + 2)2
4
+
(𝑦 + 1)2
25
= 1

11. (𝑥 + 2)2
4
+
(𝑦 + 1)2
25
= 1
Fifth step: Express in the final answer in this form
(𝑥−ℎ)2
𝑏2 +
(𝑦−𝑘)2
𝑎2 = 1
(𝑥+2)2
22 +
(𝑦+1)2
52 = 1

12. Since the equation
(𝑥+2)2
22 +
(𝑦+1)2
52 = 1 is already written in standard
form lets identify the values of a, b and c.
Note: To determine c we will use the formula 𝑐2
= 𝑎2
− 𝑏2
, 𝑐2
= 52
− 22
, 𝑐 = 21
a. Center: C(h,k) = (-2, -1)
b. vertices: 𝑉1 ℎ, 𝑘 + 𝑎 = −2, 4 𝑎𝑛𝑑 𝑉2 ℎ, 𝑘 − 𝑎 = (−2, −6)
c. foci: 𝐹1 ℎ, 𝑘 + 𝑐 = −2, −1 + 21 𝑎𝑛𝑑 𝐹2 ℎ, 𝑘 − 𝑐 = −2, −1 − 21
d. length of minor axis. 2a = 10
e. length of major axis:2b=4
g. graph the equation.

13. Find the equation of the ellipse satisfying the following conditions. Write
the equations in general form. The center is at (3, 4), a focus is at (8, 4)
and the length of the major axis is 16.

14. Find the equation of the ellipse satisfying the following conditions. Write
the equations in general form. The center is at (3, 4), a focus is at (8, 4)
and the length of the major axis is 16.

15. Solution 1: In order
to find the equation
we must plot the
center and the
focus.

16. Based from this illustration, the focus
is 5 units to the right of the center
which gives us a hint that the ellipse is
horizontal. The length of the major
axis is 16 hence a=8. We need to
determine the value of b in order to
write the equation in the form
(𝑥−ℎ)2
𝑎2 +
(𝑦−𝑘)2
𝑏2 = 1

17. The calculate for b, use
𝑐2
= 𝑎2
− 𝑏2
,
52 = 82 − 𝑏2 ,
𝑏2
= 64 − 25,
b = 39

18. Solution 2: Now we have all the values needed we will now
substitute these in the equation
(𝑥−ℎ)2
𝑎2 +
(𝑦−𝑘)2
𝑏2 = 1
(𝑥−3)2
82 +
(𝑦−4)2
39
2 = 1
First step: substitute these values in the standard form.

19. (𝑥−3)2
82 +
(𝑦−4)2
39
2 = 1
Second step: Expand the binomial and multiply both sides by the LCM
2496
(𝑥−3)(𝑥−3)
64
+
(𝑦−4)(𝑦−4)
39
= 1 2,496

20. 2496
(𝑥−3)(𝑥−3)
64
+
(𝑦−4)(𝑦−4)
39
= 1 2,496
39 𝑥2
− 6𝑥 + 9 + 64 𝑦2
− 8𝑦 + 16 = 2496
Third step: Combine common terms and simplify the result
39𝑥2
− 234𝑥 + 351 + 64𝑦2
− 512𝑦 + 1024 − 2496 = 0

21. 39𝑥2
− 234𝑥 + 351 + 64𝑦2
− 512𝑦 + 1024 − 2496 = 0
Fourth step: Express the final answer in general equation form.
39𝑥2 +64𝑦2 −234𝑥 − 512𝑦 − 1121 = 0
Therefore, equation for the ellipse that satisfies the given conditions whose center is at (3, 4), a
focus is at (8, 4) and the length of the major axis is 16 is
39𝑥2 +64𝑦2 −234𝑥 − 512𝑦 − 1121 = 0

22. A bridge is built in the shape of a semi-elliptical arch and has a
span of 200ft. The height of the arch at the distance of 60ft from
the center is 30 ft. Find the height of the arch at its center.

23. A bridge is built in the shape of a semi-elliptical arch and has a
span of 200ft. The height of the arch at the distance of 60ft from
the center is 30 ft. Find the height of the arch at its center.
Solution: To answer this problem, its best to make an illustration and
identify the given values and what is asked.

24. • To illustrate the situation,
the coordinates must be
plotted on the cartesian
plane with center at the
origin. Since the span of
the arch is 200 ft its
coordinates must be
located at (-100,0) and
(100,0). And one point
must be (60, 30).

25. Solution: The given values we have C(0,0), a point (60,30) and a=100.
(𝑥−ℎ)2
𝑎2 +
(𝑦−𝑘)2
𝑏2 = 1

26. (𝑥−ℎ)2
𝑎2 +
(𝑦−𝑘)2
𝑏2 = 1
First step: substitute these values in the standard form.
(60−0)2
1002 +
(30−0)2
𝑏2 = 1

27. (60−0)2
1002 +
(30−0)2
𝑏2 = 1
Second step: Combine common terms and simplify the result
3600
10000
+
900
𝑏2 = 1
25𝑏2 9
25
+
900
𝑏2 = 1 25𝑏2

28. 25𝑏2 9
25
+
900
𝑏2 = 1 25𝑏2
9𝑏2
+ 22500 = 25𝑏2
Third step: Solve for b
𝑏2 =
22500
16

29. 𝑏2 =
22500
16
𝒃 = 𝟑𝟕. 𝟓
The height of the arch at its center is 37.5 ft.