Conic section- Hyperbola STEM TEACH

1. CONIC SECTION

2. HYPERBOLA

3. 1. Express the equation 𝒙 𝟐 − 𝟐𝒚 𝟐 − 𝟐𝒙 − 𝟖𝒚 − 𝟐𝟕 = 𝟎 of the hyperbola
in standard form and solve for the following:
a. Center
b. Vertices
c. Foci
d. Equation of Transverse Axis
e. Length of the Transverse Axis
f. Equation of Conjugate Axis
g. Length of the Conjugate Axis
h. asymptotes

4. Solution: In order to write the equation in standard form, consider first
which coefficient (between x2 and y2) has a negative value then arrange
them in such a way that the first term must be a positive value.
𝒙 𝟐
− 𝟐𝒚 𝟐
− 𝟐𝒙 − 𝟖𝒚 − 𝟐𝟕 = 𝟎

5. 𝒙 𝟐
− 𝟐𝒚 𝟐
− 𝟐𝒙 − 𝟖𝒚 − 𝟐𝟕 = 𝟎
First Step: Arrange the order by grouping like variables
𝒙 𝟐
− 𝟐𝒙 − 𝟐𝒚 𝟐
− 𝟖𝒚 = 𝟐𝟕

6. 𝒙 𝟐
− 𝟐𝒙 − 𝟐𝒚 𝟐
− 𝟖𝒚 = 𝟐𝟕
Second Step: Factor out the coefficient
(𝒙 𝟐
−𝟐𝒙) − 𝟐(𝒚 𝟐
+ 𝟒𝒚) = 𝟐𝟕

7. (𝒙 𝟐
−𝟐𝒙) − 𝟐(𝒚 𝟐
+ 𝟒𝒚) = 𝟐𝟕
Third Step: Perform completing the square process
(𝒙 𝟐
−𝟐𝒙 + 𝟏) − 𝟐(𝒚 𝟐
+ 𝟒𝒚 + 𝟒) = 𝟐𝟕 + 𝟏 − 𝟖

8. (𝒙 𝟐
−𝟐𝒙 + 𝟏) − 𝟐(𝒚 𝟐
+ 𝟒𝒚 + 𝟒) = 𝟐𝟕 + 𝟏 − 𝟖
Fourth Step: Factor the trinomial and simplify the constant
(𝒙 − 𝟏) 𝟐
−𝟐 𝒚 + 𝟐 𝟐
= 𝟐𝟎

9. (𝒙 − 𝟏) 𝟐
−𝟐 𝒚 + 𝟐 𝟐
= 𝟐𝟎
Fifth Step: Divide both side by 20 to make the constant equal to 1
(𝒙 − 𝟏) 𝟐
−𝟐 𝒚 + 𝟐 𝟐
𝟐𝟎
=
𝟐𝟎
𝟐𝟎

10. (𝒙 − 𝟏) 𝟐
−𝟐 𝒚 + 𝟐 𝟐
𝟐𝟎
=
𝟐𝟎
𝟐𝟎
Sixth Step: Simplify the fraction, then you got the standard form.
(𝒙 − 𝟏) 𝟐
𝟐𝟎
−
𝒚 + 𝟐 𝟐
𝟏𝟎
= 𝟏

11. (𝒙 − 𝟏) 𝟐
𝟐𝟎
−
𝒚 + 𝟐 𝟐
𝟏𝟎
= 𝟏
Center (h, k) = (1, -2)
Vertices (h±a, k) 1 + 2 5, −2 , 1 − 2 5, −2
Foci (h±c, k) 1 + 30, −2 , 1 − 30, −2
Equation of the Transverse Axis: y=k, y = -2
Length of the Transverse Axis: 2a2(2 5)= 4 5
Equation of the Conjugate Axis: x=h, x = 1
Length of the Conjugate Axis: 2b, 2( 10)= 2 10

12. Asymptotes: 𝑦 = 𝑘 ±
𝑏
𝑎
𝑥 − ℎ
• 𝑦 = −2 ±
10
2 5
(𝑥 − 1)
• 𝑦 = −2 ±
1
2
10
5
(𝑥 − 1)
• 𝒚 = −𝟐 ±
𝟐
𝟐
(𝒙 − 𝟏)
Since x2, therefore the value of a is found under x2
𝑎 = 20 = 2 5 𝑏 = 10 𝑐 = 𝑎2 + 𝑏2 = 20 + 10 = 30

13. Write the equation of the hyperbola in standard and general
form that satisfies the given conditions. The center is at (7, -2), a
vertex is at (2, -2) and an endpoint of a conjugate axis is at (7, -6).

14. Write the equation of the hyperbola in standard and general
form that satisfies the given conditions. The center is at (7, -2), a
vertex is at (2, -2) and an endpoint of a conjugate axis is at (7, -6).
Solution: In order to write the equation, you only need the value for the
center, a and b. Then determine where the hyperbola is facing and the
appropriate standard form to be used.

15. The center (h, k) = (7, -2) is midway between the vertices.
The distance between the vertex and the center is |7-5|=5, hence a = 5.
The distance between the endpoint of the conjugate axis and the center is|-6 – (-2)|= 4, hence, b= 4. If a = 5 and b
= 4, then, 𝑐 = 𝑎2 + 𝑏2 = 52 + 42 = 41
The center and a vertex are both contained on the focal axis. Since they have the same y -coordinate, the focal axis
is horizontal. Therefor the, equation of the ellipse is of the form
(𝑥−ℎ)2
𝑎2 −
𝑦−𝑘 2
𝑏2 = 1

16. By substitution, the equation of the ellipse in
standard form is
(𝒙−𝟕) 𝟐
𝟐𝟓
−
𝒚+𝟐 𝟐
𝟏𝟔
= 𝟏.

17. Solution: To write it in general form
follow the following steps:
Step 1: Multiply both sides by the product of 25 and 16
𝟐𝟓(𝟏𝟔)
(𝒙−𝟕) 𝟐
𝟐𝟓
−
𝒚+𝟐 𝟐
𝟏𝟔
= 𝟏 𝟐𝟓(𝟏𝟔)

18. 𝟐𝟓(𝟏𝟔)
(𝒙−𝟕) 𝟐
𝟐𝟓
−
𝒚+𝟐 𝟐
𝟏𝟔
= 𝟏 𝟐𝟓(𝟏𝟔)
Step 2: Simplify the result
𝟏𝟔(𝒙 − 𝟕) 𝟐
−𝟐𝟓 𝒚 + 𝟐 𝟐
= 𝟒𝟎𝟎

19. 𝟏𝟔(𝒙 − 𝟕) 𝟐
−𝟐𝟓 𝒚 + 𝟐 𝟐
= 𝟒𝟎𝟎
Step 3:Expand the binomial (you may use foil method)
𝟏𝟔 𝒙 𝟐
− 𝟏𝟒𝒙 + 𝟒𝟗 − 𝟐𝟓 𝒚 𝟐
+ 𝟒𝒚 + 𝟒 = 𝟒𝟎𝟎

20. 𝟏𝟔 𝒙 𝟐
− 𝟏𝟒𝒙 + 𝟒𝟗 − 𝟐𝟓 𝒚 𝟐
+ 𝟒𝒚 + 𝟒 = 𝟒𝟎𝟎
Step 4:Distribute the multiplier to the trinomial
𝟏𝟔𝒙 𝟐
− 𝟐𝟐𝟒𝒙 + 𝟕𝟖𝟒 − 𝟐𝟓𝒚 𝟐
− 𝟏𝟎𝟎𝒚 − 𝟏𝟎𝟎 − 𝟒𝟎𝟎 = 𝟎

21. 𝟏𝟔𝒙 𝟐
− 𝟐𝟐𝟒𝒙 + 𝟕𝟖𝟒 − 𝟐𝟓𝒚 𝟐
− 𝟏𝟎𝟎𝒚 − 𝟏𝟎𝟎 − 𝟒𝟎𝟎 = 𝟎
Step 5:Combine the constant then arrange it in order
𝟏𝟔𝒙 𝟐
− 𝟐𝟓𝒚 𝟐
− 𝟐𝟐𝟒𝒙 − 𝟏𝟎𝟎𝒚 + 𝟐𝟖𝟒 = 𝟎
The standard form is
(𝒙−𝟕) 𝟐
𝟐𝟓
−
𝒚+𝟐 𝟐
𝟏𝟔
= 𝟏 and the general equation form is
𝟏𝟔𝒙 𝟐 − 𝟐𝟓𝒚 𝟐 − 𝟐𝟐𝟒𝒙 − 𝟏𝟎𝟎𝒚 + 𝟐𝟖𝟒 = 𝟎

22. An architect designs two houses that are shaped and
positioned like a part of the branches of the hyperbola
whose equation is 625𝑦2
− 400𝑥2
= 250,000where x and
y are in yards. How far apart are the houses at their closest
point?

23. An architect designs two houses that are shaped and
positioned like a part of the branches of the hyperbola
whose equation is 625𝑦2
− 400𝑥2
= 250,000where x and
y are in yards. How far apart are the houses at their closest
point?

24. Solution: The closest point of these houses are the vertices of
the hyperbola. In order to determine their distances, we need
to write the equation in standard form.

25. 625𝑦2
− 400𝑥2
= 250,000
Multiply both sides by
𝟏
250,000
𝟏
250,000
625𝑦2
− 400𝑥2
= 250,000
𝟏
250,000

26. 𝟏
250,000
625𝑦2
− 400𝑥2
= 250,000
𝟏
250,000
Simplify the result
𝒚 𝟐
𝟒𝟎𝟎
−
𝒙 𝟐
𝟔𝟐𝟓
= 𝟏

27. 𝒚 𝟐
𝟒𝟎𝟎
−
𝒙 𝟐
𝟔𝟐𝟓
= 𝟏
Determine the center and a value
𝒚 𝟐
𝟐𝟎 𝟐
−
𝒙 𝟐
𝟐𝟓 𝟐
= 𝟏

28. C(0,0), a = 20
Vertices (0, 20) and (0, -20)
The two houses are 40 yards apart.