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Conic section- Hyperbola STEM TEACH
The document provides the step-by-step solution to solving a hyperbola equation in standard form. It gives the center, vertices, foci, transverse and conjugate axes, and asymptotes. It then solves two word problems involving hyperbolas, determining the distance between houses shaped as branches of a hyperbola based on the given equation.
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Conic section- Hyperbola STEM TEACH
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- 3. 1. Express theequation 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 orderto 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 − 𝟐𝒚𝟐 − 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 equationof 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 equationof 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, theequation of the ellipse in standard form is (𝒙−𝟕) 𝟐 𝟐𝟓 − 𝒚+𝟐 𝟐 𝟏𝟔 = 𝟏.
- 17. Solution: To writeit 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 designstwo 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 designstwo 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 closestpoint 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 Multiplyboth 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 Simplifythe 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.