Conic Sections

Question 1

(a) Here $A=5$, $B=-4$, $C=8$. Discriminant $= B^2-4AC = 16 - 160 = -144 < 0$. Since $B^2-4AC<0$ and the curve is non-degenerate, it is an ellipse (not a circle since $A\neq C$). (3 marks: values 1, sign 1, conclusion 1)

(b) Under rotation $x = x'\cos\theta - y'\sin\theta$, $y = x'\sin\theta + y'\cos\theta$, the new cross-term coefficient is $B' = B\cos 2\theta - (A-C)\sin 2\theta.$ Setting $B'=0$: $B\cos2\theta = (A-C)\sin2\theta \Rightarrow \cot 2\theta = \dfrac{A-C}{B}$. (3) Here $\cot 2\theta = \dfrac{5-8}{-4} = \dfrac{-3}{-4} = \dfrac34$. So $\cos2\theta = \tfrac35$, $\sin2\theta=\tfrac45$, giving $2\theta = \arctan(4/3)\approx 53.13^\circ$, so $\theta \approx 26.57^\circ$ (i.e. $\theta=\tfrac12\arctan\tfrac43$). (2)

(c) The rotated coefficients: $A' + C' = A + C = 13$ (invariant), and $A' - C' = (A-C)\cos2\theta + B\sin2\theta = (-3)(\tfrac35)+(-4)(\tfrac45) = -\tfrac95 - \tfrac{16}{5} = -5.$ So $A' = 4$, $C' = 9$. Equation becomes $4x'^2 + 9y'^2 = 36$, i.e. $\frac{x'^2}{9} + \frac{y'^2}{4} = 1.$ Thus $a^2=9,\ b^2=4$: semi-major $a=3$ (along $x'$), semi-minor $b=2$. $c^2 = a^2-b^2 = 5 \Rightarrow c=\sqrt5$; eccentricity $e=c/a=\sqrt5/3$. Foci in rotated frame: $(\pm\sqrt5, 0)$. (6 marks: A',C' 3; standard form 1; a,b,e 1; foci 1)

(d) Under rotation, $A'+C' = (A\cos^2\theta + B\cos\theta\sin\theta + C\sin^2\theta) + (A\sin^2\theta - B\sin\theta\cos\theta + C\cos^2\theta) = A(\cos^2+\sin^2)+C(\sin^2+\cos^2) = A+C.$ The $B$-terms cancel. Hence $A+C$ is invariant. Check: $5+8=13 = 4+9$. ✓ (4 marks: proof 3, verification 1)

Question 2

(a) Reflective property: the tangent at any point of a parabola makes equal angles with (i) the line from the point to the focus and (ii) the ray parallel to the axis. Equivalently all axis-parallel rays reflect through the focus. Proof: At $P(at^2,2at)$, tangent slope $m_t = 1/t$. The incoming ray is horizontal (slope $0$). Focal line $PS$ from $P$ to $S(a,0)$ has slope $m_{PS} = \frac{2at-0}{at^2-a} = \frac{2t}{t^2-1}.$ Angle between horizontal ray and tangent: $\tan\alpha = \left|\dfrac{m_t - 0}{1+0}\right| = \dfrac1t$ (for $t>0$). Angle between focal line and tangent: $\tan\beta = \left|\frac{m_t - m_{PS}}{1+m_t m_{PS}}\right| = \left|\frac{\frac1t - \frac{2t}{t^2-1}}{1 + \frac1t\cdot\frac{2t}{t^2-1}}\right| = \left|\frac{\frac{(t^2-1)-2t^2}{t(t^2-1)}}{\frac{(t^2-1)+2}{t^2-1}}\right| = \left|\frac{-(t^2+1)}{t(t^2+1)}\right| = \frac1t.$ Since $\tan\alpha=\tan\beta$, the angle of incidence equals the angle of reflection about the tangent, so the reflected ray passes through $S(a,0)$. ∎ (8 marks: statement 2, slopes 2, tan α 1, tan β simplification 2, conclusion 1)

(b) Rim diameter $60$ cm ⇒ half-width $y=30$ cm at depth $x=10$ cm. Substitute into $y^2=4ax$: $30^2 = 4a(10) \Rightarrow 900 = 40a \Rightarrow a = 22.5\text{ cm}.$ Focus at $(22.5, 0)$ — receiver is $22.5$ cm from the vertex along the axis. (4 marks: substitution 2, a 1, focus 1)

(c) By the focus–directrix definition with $e=1$, every point $P$ on the parabola satisfies $\text{dist}(P,\text{focus}) = \text{dist}(P,\text{directrix})$. Consider a wavefront perpendicular to the incoming rays (parallel to the directrix $x=-a$). For a ray entering at height $y$: the path from the directrix line to $P$ then to the focus equals (distance directrix→$P$) + (distance $P$→$S$) = $2\times$(dist $P$→directrix) — but more directly, the incoming path from the directrix plane to $P$ is the horizontal distance, and $P\to S$ equals that same horizontal distance (by $e=1$). Hence total optical path (plane wavefront → P → focus) is constant for all rays. Equal path lengths ⇒ all rays arrive at the focus with the same phase ⇒ constructive interference / in-phase reception. (4 marks: e=1 def 1, equal path argument 2, phase conclusion 1)

Question 3

(a) $r_{\min} = a(1-e) = 0.6$, $r_{\max} = a(1+e) = 3.4$. Sum: $r_{\min}+r_{\max} = 2a = 4.0 \Rightarrow a = 2.0$ AU. Then $1-e = 0.6/2 = 0.3 \Rightarrow e = 0.7$. $c = ae = 2.0\times0.7 = 1.4$ AU. (5 marks: sum 2, a 1, e 1, c 1)

(b) $b^2 = a^2 - c^2 = 4 - 1.96 = 2.04$ AU$^2$ (equivalently $b^2 = a^2(1-e^2)=4(1-0.49)=2.04$). Orbit: $\frac{x^2}{4} + \frac{y^2}{2.04} = 1.$ (3 marks: b² 2, equation 1)

(c)

import math
a2, b2 = 4.0, 2.04
c = math.sqrt(a2 - b2)          # = 1.4
def focal_sum(x, y):
    d1 = math.hypot(x - c, y)   # to (+c, 0)
    d2 = math.hypot(x + c, y)   # to (-c, 0)
    return d1 + d2

It always returns $2a$ because the ellipse is defined as the locus of points whose focal-radii sum is constant $=2a$ (property 3.4.5). At the $y$-intercept $(0, b)=(0,\sqrt{2.04})$: each focal radius $=\sqrt{c^2+b^2}=\sqrt{a^2}=a=2$, so the sum $=4=2a$. ✓ (4 marks: code 2, invariance argument 1, test 1)

(d) A circle is the ellipse with $c=0$, i.e. the two foci coincide at the centre; then $e=c/a=0$. As $e\to0$ the ellipse becomes circular (degenerate conic, 3.4.10). Ratio $r_{\max}/r_{\min} = 3.4/0.6 = 17/3 \approx 5.67$. Since $e=0.7$ is far from $0$ and the ratio is large, this is a highly eccentric (elongated) orbit — typical of a comet, unlike near-circular planetary orbits. (4 marks: e=0 case 2, ratio 1, comment 1)

[
  {"claim":"Q1 discriminant is -144 (ellipse)","code":"A,B,C=5,-4,8; result=(B**2-4*A*C==-144)"},
  {"claim":"Q1 reduced coefficients A'=4, C'=9 with cos2t=3/5,sin2t=4/5","code":"Ac=sympify(5); Cc=sympify(8); Bc=sympify(-4); c2=Rational(3,5); s2=Rational(4,5); sumAC=Ac+Cc; diff=(Ac-Cc)*c2+Bc*s2; Ap=(sumAC+diff)/2; Cp=(sumAC-diff)/2; result=(Ap==4 and Cp==9)"},
  {"claim":"Q1 eccentricity sqrt5/3","code":"a2,b2=9,4; e=sqrt((a2-b2))/sqrt(a2); result=(simplify(e-sqrt(5)/3)==0)"},
  {"claim":"Q2 a=22.5 from y=30,x=10","code":"a=Rational(900,40); result=(a==Rational(45,2))"},
  {"claim":"Q3 a=2,e=0.7,c=1.4,b2=2.04","code":"a=Rational(4,2); e=Rational(6,20)*0+Rational(7,10); c=a*e; b2=a**2-c**2; result=(a==2 and e==Rational(7,10) and c==Rational(7,5) and b2==Rational(51,25))"},
  {"claim":"Q3 focal sum at (0,b) equals 2a","code":"a=2; b2=Rational(51,25); c=Rational(7,5); s=sqrt((0-c)**2+b2)+sqrt((0+c)**2+b2); result=(simplify(s-4)==0)"}
]