3. Mind Warm Ups - Conics

Mind-Warm-Ups Conics: The Ellipse

Problem 1: what are conic sections or conics?
Ans: consider the double cone of figure 1. Some of the cross sections (plane slices) that can be obtained from the double cone are:
(a) Single point
(b) Single line
(c) Double lines
(d) Circle
(e) Parabola
(f) Ellipse
(g) Hyperbola
These shapes are called conic sections or conics. It is shown in problem 5 that the circle is a special case of an ellipse. The parabola, ellipse and hyperbola are called non-degenerate conics while the single point, single line and double lines are called degenerate conics

Problem 2: the standard forms of the equations of non-degenerate conics are different because the values of their respective eccentricities are different. What is meant by the eccentricity of a non-degenerate conic?
Ans: The eccentricity of a non-degenerate conic is a constant (denoted by e) that is incorporated into its definition. In general, non-degenerate conics are defined as the set of points P in R2 that satisfy the following distance equality:
       PF = ePD -------(1).
Where PF is the distance from P to a fixed point F called the focus of the conic; PD is the distance from P to a fixed line called the directrix and e (the eccentricity) is a constant multiple of PD.
A non-degenerate conic is an ellipse if 0 ≤ e < 1 (a circle if e = 0), a parabola if e = 1, a hyperbola if e > 1.

Problem 3: consider the ellipse in figure 2. The point O(0,0) is the origin of the y and x axes. The ellipse has two foci: F(ae,0) and F'(-ae,0), and two directrixes. The first directrix MN, is the line x = a/e. The second directrix M'N' is the line x =-a/e. P(x,y) is an arbitrary point on the ellipse.
Determine the standard form of the equation of the ellipse in figure 2 if PD is a perpendicular line from P to the directrix MN and PF is the line from P to the focus, F.
Ans: (a) PF = ePD
Therefore PF2 = e2PD2
Therefore (x - ae)2 + y2 = e2(x - a/e)2 = (ex - a)2.
Therefore x2 - 2aex + a2e2 + y2 = e2x2 - 2aex + a2.
Therefore x2(1 - e2) + y2 = a2(1 - e2)
Therefore x2/a2 + y2/[a2(1 - e2)] = 1
If we let b = a√(1 - e2), then b2 = a2(1 - e2).
Therefore x2/a2 + y2/b2 = 1 -------(2).
Equation (2) is the standard form of the equation of an ellipse. This equation is symmetrical in x and y, hence the ellipse has two foci and two directrixes. The ellipse intersects the x axis at the points(a,0) and (-a,0) and intersects the y axis at the points (b,0) and (-b,0). The lines joining the points of intersection on the x axis is called the major axis of the ellipse. The line joining the points of intersections on the y axis is called the minor axis of the ellipse since a > b.

Problem 4 : what parametric equations can describe the ellipse x2/a2 + y2/b2 = 1?
Ans: x = acost; y = asint.      - π < t ≤ π, or 0 < t ≤ 2π.

Problem 5 : show that the equation of an ellipse in standard form is the equation of a circle when e = 0.
Ans: the equation of an ellipse in standard form is:
    x2/a2 + y2/b2 = 1
Recall that b2 = a2(1 - e2). When e = 0, b =a.
Therefore     x2/b2 + y2/b2 = 1
Therefore     x2 + y2 = b2 which is the equation of a circle with center at the origin and radius b units

Problem 6: an ellipse E in standard form has the following equation:
    x2/a2 + y2/b2 = 1
(a) Determine the equation of the tangent to E at the point with parameter t.
(b) Determine the equation of the normal to the tangent to E at the point with parameter t = π/4 when a =3 and b =1.
Ans: parametric equations of E: y = bsint; x = acost      -π < t ≤ π
(a) The gradient of the tangent to E at the point with parameter t = (dy/dt)(dt/dx).
Therefore gradient = bcost/-asint .
Therefore (y - bsint)/(x - acost) = bcost/-asint.
Therefore y - bsint = -bcost(x - acost)/asint
Therefore xbcost + yasint = abcos2t + absin2t = ab(cos2t + basin2t) = ab
Dividing both sides by ab gives:
xcost/a + ysint/b = 1. This is the equation of the tangent to E at parameter t.
xx1/a2 + yy1/b2 = 1 is the equation of the tangent to E at point (x1,y1)
(b) Gradient of normal to tangent = -/gradient of tangent.
Therefore gradient of normal = asint/bcost.
Therefore (y -bsint)/x -acost) = asint/bcost.
Therefore ybcost -axsint = b2sintcost - a2sintcost.
For a =3, b = 1 and t = π/4, we have:
y = 3x -4√2, as the equation of the normal to the tangent to E at parameter t.

Problem 7: Prove that light which comes from one focus of an elliptic mirror E (figure 3a), is reflected at the ellipse to pass through the second focus.
Ans: objective: prove that <TPF = <QPF'.
consider figure 3b. QT is the tangent to the ellipse at P and it intersects the x axis at T. Its parametric equation is:
xcost/a + ysint/b = 1
Therefore, T is the point (a/cost,0).
Therefore TF = (a/cost) - ae; TF' = (a/cost) + ae
Therefore TF/TF' = ((a/cost) - ae)/((a/cost) + ae) = (1 - ecost)/(1 + ecost)
PF = ePD (definition of an ellipse)
Therefore PF = e(a/e - acost) = a - aecost.
PF' = ePD'
Therefore PF' = e(a/e + acost) = a + aecost.
Therefore PF/PF' = (a - aecost)/(a + aecost) = (1 - ecost)/(1 + ecost)
Therefore PF/PF' = TF/TF'.
Therefore PF/TF = PF'/TF'
If we apply the sine rule to triangle PFT and triangle PF'T, we get
PF/TF = sin<PTF/sin<TPF and PF'/TF' = sin<PTF'/sin<TPF'
Therefore sin<PTF/sin<TPF = sin<PTF'/sin<TPF'
   <PTF = sin<PTF' (same angle).
Therefore sin<TPF = sin<TPF'.
Since <TPF ≠ <TPF', it follows that
    <TPF = π - <TPF'.
Therefore <TPF = <SPT = <QPF' (vertically opposite).
Therefore angle of incidence of light ray = angle of reflection.
Therefore the reflected light ray passes through the second focus F'(ae,0).

Problem 8: Consider the following equation:
   9x2 + 16y2 = 144.
(a) Is this equation an ellipse?
(b) If the equation is an ellipse what are the lengths of its major axis and minor axis?
(c) What is its eccentricity?
(d) Where are its foci?
Ans: (a) divide the equation by 144 to obtain:
x2/16 + y2/9 = 1
= x2/42 + y2/32 = 1
Therefore equation is an ellipse with a = 4 and b = 3.
(b) Length of major axis = 2a = 8. Length of minor axis = 2b = 6
(c) b2 = a2(1 - e2)
Therefore 16e2 = 7. Therefore e = √(7/16) = √7/4.
(d)First focus F is at point (ae,0) = (√7,0). Second focus F' is at (-ae,0) = (-√7,0).

Problem 9: consider the following equation:
   25x2 + 16y2 = 400
Along which lines are the major and minor axes of the ellipse represented by this equation?
Ans: the major axis is along the Y axis and the minor axis is along the X axis.

Problem 10: The first of Kepler's Laws of Planetary Motions states that the orbit of a planet is an ellipse, with the sun at a focus.
Suppose that the elliptical orbit of a planet is represented by the following equation of an ellipse in standard form:
   25x2 + 64y2 = 1600
Determine the distance of the planet from the sun when the planet is at the point where the ellipse intersects the y axis.
Ans: No help here :).

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Peter Oye Sagay

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