﻿3. Mind Warm Ups - Conics

## Mind-Warm-Ups Conics: The Hyperbola

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 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 hyperbola in figure 2a. The point O(0,0) is the origin of the y and x axes. The hyperbola 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 hyperbola.
Determine the standard form of the equation of the hyperbola 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.
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 a2(e2 - 1) = x2(e2 - 1) - y2.
Therefore x2/a2 - y2/[a2(e2 - 1)] = 1
If we let b = a√(e2 - 1), then b2 = a2(e2 - 1).
Therefore x2/a2 - y2/b2 = 1 -------(2).
Equation (2) is the standard form of the equation of a hyperbola. This equation is symmetrical in x and y, hence the hyperbola has two foci and two directrixes. The hyperbola 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 line joining the points of intersection on the x axis is called the major axis or transverse axis of the hyperbola. The line joining the points of intersections on the y axis is called the minor axis or conjugate axis of the hyperbola. The lines y = (b/a)x and y = -(b/a)x (figure 2b) are called the asymptotes of the hyperbola. They are derived by setting
x2/a2 - y2/b2 = 0. As x gets infinitely large in either direction, the hyperbola tends to the asymptotes (the two never meet).

Problem 4 : what parametric equations can describe the hyperbola x2/a2 - y2/b2 = 1?
Ans: x = asect; y = btant.      - π/2 < t < π/2 U π/2 < t < 3π/2.

Problem 5 : Determine the asymptotes of the following hyperbola:
y = 1/(1 -x)
Ans: x = 1; y = 0.

Problem 6: a hyperbola E has the following equation:
x2 - 2y2 = 1
(a) Determine the lengths of the tranverse and conjugate axes.
(b) Determine the foci of E.
(c) Determine the gradient of FP and F'P (lines not parallel to the y axis), where P is the parametric point (sect, (1/√2)tant). - π/2 < t < π/2 U π/2 < t < 3π/2.
(d) Determine the point t in the first quadrant on E for which FP is perpendicular to F'P.
Ans: parametric equations of E: y = btant; x = asect      - π/2 < t < π/2 U &pi/2; < t < 3π/2.
(a) a = 1, b = √(1/2). Therefore:
Length of tranverse axis (major axis) = 2a = 2. Length of conjugate axis (minor axis) = 2b = √2.
(b) b2 = a2(e2 - 1). Therefore, 1/2 = 1(e2 - 1)
Therefore, e = √(3/2). Therefore Foci F and F' at points (√(3/2),0) and (-√(3/2), 0) respectively.
(c) gradient of FP = [(1/√2)tant - 0]/[sect - √(3/2)] = tant/((√2)sect - √3)
gradient of F'P = [(1/√2)tant - 0]/[sect + √(3/2)] = tant/((√2)sect + √3)
(d) If FP is perpendicular to F'P, then the product of their gradients = -1.
Therefore [(1/√2)tant - 0]/[sect - √(3/2)]x[(1/√2)tant - 0]/[sect + √(3/2)] = -1
Therefore tant = 1/√3 and sect t = 2/√3
Therfore the point t in the first quadrant on E for which FP is perpendicular to F'P
= (a(2/√3),b(1/√3)) = ((2/√3),(1/√2)(1/√3)) = ((2/√3),(1/√6))

Problem 7: Given that the asymptotes of a hyperbola are mutually perpendicular:
(a) Determine the lengths of the transverse and conjugate axes of the hyperbola.
(b) Determine the equations of the asymptotes
(c) Determine the eccentricity of the hyperbola
Ans: (a)equations of asymptotes of hyperbola in standard form are:
y = (b/a)x, y = -(b/a)x. If they are mutually perpendicular then:
(b/a)(-b/a) = -1. Therefore b = a.
Both axes are equal with lengths 2a or 2b.
(b) Since b = a, equations of the asymptotes are: y = x and y = -x.
(c) b2 = a2(e2 - 1). Therefore e = √2.
A hyperbola with mutually perpendicular asymptotes is called a rectangular hyperbola.

Problem 8: suppose a rectangular hyperbola is expressed as:
xy = c2, where c is some positive number.
(a) Identify the x and y axes of the rectangular hyperbola
(b) write a parametric equation for the rectangular hyperbola for t ≠ 0.
Ans: (a) The x and y axes are the mutually perpendicular asymptotes.
(b) x = ct, and y = c/t.

Problem 9: consider figure 3. Show that |PF' - PF| is a constant for all points P on the hyperbola with major axis (-a,a) and foci F and F', where P is a point on the branch of the hyperbola closer to F. Lines MN and M'N' are the directrixes of the hyperbola.
Ans: PF = ePD and PF' = ePD'.
Therefore, PF' - PF = eDD'.
Therefore, PF' - PF = 2a (directrixes are at x = a/e and x = -a/e)
Similarly, PF' - PF = -2a , when P is on the branch with F' as focus.
Therefore |PF' - PF| = 2a whis is a constant.

Problem 10:(a) the reflection property of non-degenerate conics can be summarized thus:
All mirrors in the shape of a non-degerate conic reflect light coming from or going to one focus towards the other focus. Where is the other focus of the parabola?
(c) Show that the polar form of the equation of a conic can be expressed as:
r = l/(1 + ecosθ). Where l is a constant.

Ans: No help here :).

Peter Oye Sagay