CONIC SECTION
HYPERBOLA
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
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
𝒙 𝟐
− 𝟐𝒚 𝟐
− 𝟐𝒙 − 𝟖𝒚 − 𝟐𝟕 = 𝟎
First Step: Arrange the order by grouping like variables
𝒙 𝟐
− 𝟐𝒙 − 𝟐𝒚 𝟐
− 𝟖𝒚 = 𝟐𝟕
𝒙 𝟐
− 𝟐𝒙 − 𝟐𝒚 𝟐
− 𝟖𝒚 = 𝟐𝟕
Second Step: Factor out the coefficient
(𝒙 𝟐
−𝟐𝒙) − 𝟐(𝒚 𝟐
+ 𝟒𝒚) = 𝟐𝟕
(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
(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
(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
(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
(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
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
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).
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.
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
By substitution, the equation of the ellipse in
standard form is
(𝒙−𝟕) 𝟐
𝟐𝟓
−
𝒚+𝟐 𝟐
𝟏𝟔
= 𝟏.
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)
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
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
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
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
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?
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?
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.
625𝑦2
− 400𝑥2
= 250,000
Multiply both sides by
𝟏
250,000
𝟏
250,000
625𝑦2
− 400𝑥2
= 250,000
𝟏
250,000
𝟏
250,000
625𝑦2
− 400𝑥2
= 250,000
𝟏
250,000
Simplify the result
𝒚 𝟐
𝟒𝟎𝟎
−
𝒙 𝟐
𝟔𝟐𝟓
= 𝟏
𝒚 𝟐
𝟒𝟎𝟎
−
𝒙 𝟐
𝟔𝟐𝟓
= 𝟏
Determine the center and a value
𝒚 𝟐
𝟐𝟎 𝟐
−
𝒙 𝟐
𝟐𝟓 𝟐
= 𝟏
C(0,0), a = 20
Vertices (0, 20) and (0, -20)
The two houses are 40 yards apart.

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Conic section- Hyperbola STEM TEACH

  • 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.