Mathematics, 08.10.2020 08:01, sindy35111

* Anyone please give me some information about the following; • About cone and conic section.(with diagram)
• Describe how are cones, ecllipse, parabola, hyperbola are formed?
(I need full information as much as possible)
Thank you!

Answers: 3

Show answers

Answers

answered: jociymartinez6214

see below

Step-by-step explanation:

Types of Conics are: Circles, Ellipses, & Hyperbolas.

Here is information about each one:

Circle: Two parabolas facing each other. Distance from the center to the vertices are exactly the same.

(x - h)² + (y - k)² = r²

(h, k) is the center of the circler is the radius of the circleThe 4 vertices are: (h, k+r), (h, k-r), (h + r, k), & (h - r, k)

Ellipse: Two parabolas facing each other. Distance from the center to the vertices are not the same.

(x - h)²/a² + (y - k)²/b² = 1

(h, k) is the center of the ellipsea is the horizontal distance from the center to the vertices/co-verticesb is the vertical distance from the center to the co-vertices/verticesIf a > b, then vertices are: (h+a, k), & (h-a, k) and co-vertices are (h, k+b), (h, k-b)If b > a, then the vertices and co-vertices are reversed Use |a² - b²| = c² to find the distance from the center to the foci. If a > b, then Foci = (h+c, k) & (h-c, k). If b > a, then ± from the k-value

Example: (x - 1)²/9 + (y - 2)²/25 = 1

Center (h, k) = (1, 2) a = √9 = 3 b = √25 = 5

Vertices (b > a): (1, 2+5) & (1, 2-5) ==> (1, 7) & (1, -3)

Co-vertices: (1+3, 2) & (1-3, 2) ==> (4, 2) & (-2, 2)

Foci (b > a): c = √(25-9) = 4 --> (1, 2+4) & (1, 2-4) ==> (1, 6) & (1, -2)

Hyperbola: Two parabolas facing AWAY from each other. Distance from the center to the vertices may or may not be the same.

(x - h)²/a² - (y - k)²/b² = 1 or (y - k)²/b² - (x - h)²/a² = 1

↓ ↓

opens left and right opens up and down

(h, k) is the center of the hyperbolaIf x²-y², then vertices are: (h+a, k), & (h-a, k). There are NO Co-vertices.If y²-x², then vertices are (h, k+b), (h, k-b). There are NO Co-vertices.Use a² + b² = c² to find the distance from the center to the foci. If x²-y², then Foci = (h+c, k) & (h-c, k). If y²-x², then ±c from the k-valueThe slope of the asymptotes are ± b/a and pass through the center (h, k)

Example: (x - 1)²/9 - (y - 2)²/16 = 1

Center (h, k) = (1, 2) a = √9 = 3 b = √16 = 4

Vertices (x²-y²): (1+3, 2) & (1-3, 2) ==> (4, 2) & (-2, 2)

Foci (x²-y²): c = √(9 + 16) = 5 --> (1+5, 2) & (1-5, 2) ==> (6, 2) & (-4, 2)

Asymptotes: m = ± 4/3 through (h, k) = (1, 2) ==> y = ± 4/3(x - 1) + 2