Conic sections #2 - Questions and Answers

Question 1 What is the eccentricity of a conic section? How can you classify conic sections by eccentricity? How does eccentricity change the shape of ellipses and hyperbolas? Answer Step 1 Any conic section may be described as the collection of points with constant ratios between their distances to the focus and directrix. The eccentricity of the conic section, sometimes abbreviated as e, is the name given to that ratio. Step 2 An ellipse that is not a circle has an eccentricity (e) that is larger than zero but less than 1. If e = 1 then it is Parabola. If e > 1 then comes a hyperbola [A hyperbola's eccentricity has no upper limit and can be any real value larger than 1. A rectangular hyperbola's eccentricity √20. If e = 0 suddenly Circle appears. Step 3 An ellipse's eccentricity is almost zero if it is almost circular. An ellipse has a high degree of ovalness if its eccentricity is close to one. A conic section's eccentricity indicates how near it comes to having a circular form. The form resembles a circle less and less when the eccentricity of a conic section deviates from 0. A compressed circle resembles an ellipse. The closest analog to a circle among the other two conic forms is this one. A parabola would be the next closest shape to a circle, while a hyperbola would be the furthest, according to same reasoning. Question 2 Eliminate the parameter and obtain the standard form of the rectangular equation. Circle: x = h + r cos(θ), y = k+ r sin(θ) Use your result to find a set of parametric equations for the line or conic section. (When 0 ≤ θ < 2π.) Circle: center: (6, 3), radius: 7 Answer Step 1: Given x = h + r cos θ, y = k + sin θ Step 2: Solution x = h + rcos θ, y = k + sin θ x - h = r cosθ, y - k = r sin θ Squaring and adding (x-h)² + (y - k)² = r² sin²θ + r² cos²θ (x-h)² + (y - k)² = r² With (6,3), radius = 7 x = 6 + 7 cos θ y = 3 + 7 sin θ Question 3 Find the vertices and foci of the conic section. x²/144 - y²/324 = 36 Answer 1 Step 1 To find the vertices and foci of the conic section. Given the equation, x²/144 - y²/324 = 36 x²/144 * 1/36 - y²/324 * 1/36 = 1 x²/ 5184 - y²/ 11664= 1 Hence the given equation is form of hyperbola. Here, a² = 5184 => a = 72 b² = 11664 => b = 108 Step 2 equation(1) compare this below equation (x-h)² / a² - (y-k)² / b² = 1 c² = a² + b² c = √(a² + b²) c = √5184 + 11664 c = √16848 Vertices of hyperbola = (h, k) = (0,0) Foci of hyperbola = (±c, 0) = (±√16848, 0) Answer 2 Step 1 To find the vertices and foci of the conic section. Given the equation, x²/144 - y²/324 = 36 x²/5184 - y²/11664 = 1 Hence, the given equation is form of hyperbola. Here, a² = 5184 a = 72 b² = 11664 b = 108 Step 2 equation (1) compare this below equation (x-h)² / a² - (y-k)² / b² = 1 c² = a² + b² c = √(a² + b²) c = √5184+11664 c = √16848 Vertices of hyperbola = (h,k) = (0,0) Foci of hyperbola = (± c, 0) = (±√16848,0) Question 4 Find an equation of the conic described. Graph the equation. Ellipse; center at (0,0); focus at (0,3); vertex at (0, 5) Answer 1 Step 1: Consider the given information about comic Ellipse. Center: (0,0) Focus: (0,3) Vertex: (0,5) The major axis for the ellipse is y and minor axis is x that is, a > b. So, the equation of the ellipse is, (x-h)² / b² + (y-k)² / a² = 1 Where, (h, k) is the center of the ellipse. Compare the coordinate of vertex with (0, ±a). a = 5 Compare the coordinate of focus with (0, ±c) where, c² = a²-b². c = 3 Substitute c = 3 and a = 5 into c² = a²-b². 3² = 5²-b² b² = 25-9 b = √16 b = 4 Step 2: Substitute the value center, a and b into equation of ellipse. x²/16 + y²/25 = 1 Therefore, the equation of the ellipse is x²/16 + y²/25 = 1 Answer 2 Step 1 Consider the given information about conic Ellipse. Center: (0,0) Focus: (0,3) Vertex: (0,5) The major axis for the ellipse is y and minor axis is x that is, a>b So, the equation of the ellipse is, (x-h)² / b² + (y-k)² / a² = 1 Where, (h,k) is the center of the ellipse Compare the coordinate of vertex with (0, ±a) a=5 Compare the coordinate of focus with (0, ±c) where, c² = a²-b² c=3 Substitute c=3 and a=5 into c² = a² - b² 3² = 5²-b² b² = 25-9 b = √16 = 4 Step 2 Substitute the value center, a and b into equation of ellipse x²/16 + y²/25 = 1 Therefore, the equation of the ellipse is x²/16 + y²/25 = 1 Question 5 a) To simplify: The radical expression √20 Answer 1 Step 1 a) The given expression is √20 √20 = √(225) = √(2²5) By the property of the radicals, √ab = √a * √b, √(2²5) = √2² * √5 By the property of the radicals, √aⁿ = |a|; for even n. = 2 * √5 Hence the simplified form radical is, 2√5 Step 2 b) The given expression is 128^(1/4) Factor the above term, 128 = 2222222, 128^(1/4) = (2222222)^(1/4) = (2⁷)^(1/4) By the property of the exponents, (a^m)^n = a^(mn) = 2^(7/4) = 2^(6/4) * 2^(1/4) = 2^(6/4) * 2^(1/4) = 2^(61/4) * 2^(1/4) By the property of the radicals, ⁿ√ab = ⁿ√a * ⁿ√b, = (2^6)^(1/4) * 2^(1/4) = 2^(6/4) * 2^(1/4) = 2^(3/2) * 2^(1/4) = 2√2 2^(1/4) = 4√2 Hence, the simplified form of radical is, 4√2 Answer 2 a) Simplify the following: √20 Factor 20 under the radical sign. √20 = √4 x 5 = √2² x 5: √2²5 Simplify powers. √2²√5 = √2²√5 = 2√5 = 2√5: Answer: 2√5 b) Simplify the following: 128^(1/4) Simplify radicals. 128^(1/4) = (2⁷)^(1/4) = 2^(7/4) = 2^(4/4)2^(3/4) = 2 2^(3/4) Evaluate 2^(3/2). 2^(3/2) = 2√2 : Answer: 4√2