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Academic Skills Advice Conics The main shapes known as conics are: Parabola (also known as quadratics as looked at in the last lesson) Ellipse (including the circle which is a special case) Hyperbola (including the rectangular hyperbola which is a special case) Some of them include a π¦ 2 term as well as (or instead of) an π₯ 2 term You can recognise a conic from its general equation. The general equation tells you enough about the shape to enable you to sketch it. The general equations are as follows: (π β π)π + (π β π)π = ππ Circle: Ellipse: Hyperbola: π₯2 π¦2 π π2 + 2 π₯2 π¦2 π π2 β 2 =1 centre = (π,π) and radius= π π₯-intercepts = ±π π¦-intercepts = ±π =1 Points to note: Circle: If π and π are both 0, the equation becomes π₯ 2 + π¦ 2 = π 2 (i.e. a circle with centre at (0, 0) and radius r.) To find the centre and the radius, the coefficients of π₯ 2 and π¦ 2 must = 1. Ellipse: The equation must = 1. (i.e. you need 1 on the right hand side). You need squared numbers on the bottom of each fraction. The squared number beneath π₯ tells us the π₯-intercepts and the squared number beneath π¦ tells us the π¦-intercepts. Hyperbola: The equation must = 1. (i.e. you need 1 on the right hand side). You need squared numbers on the bottom of each fraction. The squared number beneath π₯ tells us where to draw the vertical edges of the rectangle and the squared number beneath π¦ tells us where to draw the horizontal edges of the rectangle. © H Jackson 2011 / ACADEMIC SKILLS 1 Circle Notice that for a circle: ο· the π₯ and π¦ terms are both squared, then added. ο· the coefficients of π₯ 2 and π¦ 2 are equal ο· there is no π₯π¦ term. (π β π)π + (π β π)π = ππ centre = (π,π) and radius= π Examples: ο· Sketch the circle with equation (π β π)π + (π + π)π = ππ From the equation we see that the centre of the circle is at (3, β2) and the radius is 4 (because β16 = 4) π¦ Notice that the co-ordinates of the centre have the opposite signs to the numbers in the brackets 2 π₯ -1 x (3, β2) ο· Sketch the circle with equation πππ + πππ = ππ We need the coefficients of π₯ 2 and π¦ 2 to equal 1 so we divide the equation by 3. Starting equation: ÷3: We now have: 3π₯ 2 + 3π¦ 2 = 12 3π₯ 2 3 + 3π¦ 2 3 = 12 3 π π + ππ = π Now we can see that the centre of the circle is at (0, 0) and the radius is 2 (ππ β4 = 2) π¦ 2 -2 2 π₯ -2 © H Jackson 2011 / ACADEMIC SKILLS 2 Ellipse π₯2 π2 + π¦2 π2 Notice that for an ellipse: ο· the π₯ and π¦ terms are both squared, and then added ο· there is no π₯π¦ term. ο· the ellipse intercepts the π₯-axis at ±π and the π¦-axis at ±π =1 (n.b. when π = π we have a circle). Examples: ο· Sketch the ellipse with equation ππ π ππ + ππ = π We need squared numbers on the bottom of each fraction so we letβs rewrite the equation as: π₯2 π¦2 + =1 32 52 Itβs now easy to see that the ellipse crosses the π₯-axis at ±3 and the π¦-axis at ±5 π¦ 5 π₯ 3 β3 β5 ο· Sketch the ellipse with equation πππ + πππ = ππ The equation has to equal 1 so we divide the whole equation by 32. 2π₯ 2 32 We now have: ππ + + ππ 8π¦ 2 32 ππ π 32 = 32 =π (we can rewrite this as: π₯2 42 π¦2 + 22 = 1) From the equation we see that the ellipse crosses the π₯-axis at ±4 and the π¦-axis at ±2 π¦ 2 β4 4 π₯ β2 © H Jackson 2011 / ACADEMIC SKILLS 3 Hyperbola π₯ 2 π¦2 β =1 π2 π 2 Notice that for a hyperbola: ο· the π₯ and π¦ terms are both squared and then subtracted. ο· the diagonals of the rectangle (formed with length from βπ to π and height from βπ to π) are asymptotes for the hyperbola. Examples: ο· Sketch the hyperbola with equation We can rewrite the equation as: π₯2 ππ π β ππ π =π π¦2 β 32 = 1 22 From the equation we see that we need to draw a rectangle from ±2 along the π₯-axis and from ±3 along the π¦-axis. We then use the diagonals of the rectangle as asymptotes for the hyperbola. (It might be a good idea to draw the rectangle and itβs diagonals in pencil so you can erase them when you have used them as a guide to draw your hyperbola). π¦ 3 2 π₯ Form a rectangle. ο· Use the diagonals as asymptotes. Sketch the hyperbola with equation πππ β πππ = π The equation has to equal 1 so we need to divide the whole equation by 8. 8π₯ 2 8 We now have: β π₯2 β 2π¦ 2 8 π¦2 4 8 =8 =1 We can rewrite this as: π₯2 12 π¦2 β 22 = 1 This time draw your rectangle from ±1 along the π₯-axis and from ±2 along the π¦-axis. π¦ 2 1 π₯ © H Jackson 2011 / ACADEMIC SKILLS 4