﻿ Mind Warm Ups - Conics

## Mind-Warm-Ups Conics: The Parabola

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 will be shown later in the ellipse problems 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 parabolas in figure 2. The point O(0,0) is the origin of the y and x axes. The point F(a,0) is the focus of the parabola in figure 2a. The line MN through (-a,0) is its directrix and P(x,y) is an arbitrary point on the parabola. The line containing the focus (in this case the x axis) is called the axis of the parabola. The point F(0,a) is the focus of the parabola in figure 2b. The line MN through (0,-a) is its directrix and P(x,y) is an arbitrary point on the parabola.
(a) Determine the standard form of the equation of the parabola in figure 2a if PD is a perpendicular line from P to the directrix MN and PF is the line from P to the focus, F.
(b) Determine the standard form of the equation of the parabola in figure 2b 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 = PD since e = 1 for a parabola (from equation 1 in problem 2).
Therefore PF2 = PD2
Therefore (x - a)2 + y2 = (x + a)2
Therefore x2 - 2ax + a2 + y2 = x2 + 2ax + a2
Therefore y2 = 4ax -------(2).
Equation (2) is the standard form of a parabola with focus (a,0) and directrix x = -a.
(b) PF = PD since e = 1 for a parabola (from equation 1 in problem 2)
Therefore PF2 = PD2
Therefore (y - a)2 + x2 = (y + a)2
Therefore y2 - 2ay + a2 + x2 = y2 + 2ay + a2
Therefore x2 = 4ay -------(3).
Equation (3) is the standard form of a parabola with focus (0,a) and directrix y = -a.

Problem 4 : what parametric equations can describe the parabola y2 = 4ax?
Ans: x = at2; y = 2at. t in R2.

Problem 5 : a parabola E in standard form has the following equation:
y2 = 8x
(a) Determine the focus, vertex, axis and directrix of E.
(b) Determine the equation of the chord PQ, if P and Q are distinct points on E with parameters t1 and t2 respectively.
(c) What must be the relationship between t1 and t2 for PQ to be parallel to the y axis?
Ans: (a) since 4a = 8; a = 2.
Therefore, the focus of E is (2,0), its vertex is (0,0), its axis is the x axis and the equation of its directrix is x = -2.
(b) The parametric coordinates of points P and Q are (2t12,4t1) and (2t22,4t2) respectively.
Therefore for t12 ≠ t22, the gradient of PQ is as follows:
(4t1 - 4t2)/(2t12 - 2t22) = 2(t1 - t2)/(t12 - t22) = 2/(t1 + t2)
Therefore, (y - 4t1)/(x - 2t12) = 2/(t1 + t2)
Therefore, y(t1 + t2) = 2(x - 2t12) + 4t1(t1 + t2); is the equation of the chord PQ.
(c) t1 = -t2.

Problem 6: a parabola E in standard form has the following equation:
y2 = 8x
(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 = 2.
Ans:since 4a = 8; a = 2.
parametric equations: y = 2at = 4y; x = at2
the gradient of the tangent to E at the point with parameter t = (dy/dt)(dt/dx).
Therefore, gradient = 2a/(2a)t = 1/t.
Therefore,(y - 2at1)/(x - at12) = 1/t1.
Therefore,(y - 4t1)/(x - 2t12) = 1/t1
Therefore,the equation of the tangent to E at the point with parameter t is:
yt1 = x + 2t12
or, yy1 = 2a(x + x1) -------(4)
Equation (4) is the general equation of the tangent at the point (x1,y1) to a parabola with focus at the point (a,0) and directrix at x = -a.
(b) Since the normal is perpendicular to the tangent at t, its gradient = -t
Therefore,(y - 4t1)/(x - 2t12) = -t1
Therefore equation of normal to parabola at parameter t = 2 (i.e point (8,8) ) is:
(y - 4(2))/(x - 2(2)2) = -2
That is, y = -2x + 24.

Problem 7: consider the parabola E of figure 3a. RP is a light ray parallel to the axis of E. P(at2,2at) is RP's point of incidence on the parabolic morror E. QN is a tangent to E at P. Prove that the reflected ray PF passes through the focus F(a,0) of E.
Ans: objective: to prove that <QPR = <RPF.
consider triangle PNF (N is the point where the tangent to E at P intersects the x axis).
Now, since QN is the tangent to E at P, its equation is:
ty = x + at2 (see problem 6). So, N is at (-at2,0).
Therefore (considering triangle PNF) NF = NO + OF = at2 + a
and, by the distance formula,
FP = √[(a - at2)2 + (2at)2] = √( a2 + 2a2t2 + a2t4) = at2 + a.
Therefore, NF = FP. So, triangle PNF is isosceles and so, <PNF = <NPF.
Therefore <QPR = <NPF since <PNF = <QPR (corresponding angles).
Therefore, PF is the reflected ray of the light ray RP since the angle of incidence equals the angle of reflection and it passes through F(a,0), the focus of E.

Problem 8: suppose a light source is at the focus F(a,0) of the parabolic mirror E of figure 3a problem 7 (i.e, light rays incident on E are comming from the focus of E). In what direction will the rays be reflected?
Ans: the reflected rays will be parallel to the axis of E.

Problem 9: consider the following equation in R2:
y = x2 - 12x -------(5)
Determine the curve with this equation without plotting. If this curve is a parabola where is its vertex and its axis relative to the X and Y axes?
Ans: objective is to see if equation (5) can be manipulated to represent the equation of a familiar curve without breaking any mathematical rules.
So lets add 36 to both sides of equation (5):
y + 36 = x2 - 12x + 36
Therefore, y + 36 = (x - 6)2
By setting y' = y + 36 and x' = (x - 6)2, we can write:
y' = x'2. This is the equation of a parabola with vertex (0,0) in a X' - Y'.
Therefore, relative to the X, Y axes, the equation is a parabola with vertex at (6,-36) and axis is the line x = 6.
plot to confirm.

Problem 10: Mathematicians Descartes and Fermat are credited with the discovery of coordinate geometry. Descartes was also a great philosopher. The following quote is from Descartes:
"In order to seek truth it is necessary once in the course of our life to doubt as far as possible all things."
Descartes discovered the truth of coordinate geometry. What is the essence of this truth?
Ans: No help here :).

Peter Oye Sagay