Conic Sections

Question 1 (10 marks)

Let $P(x,y)$ be a point on the conic. By focus–directrix definition with $e = 1$ (parabola): $\text{distance to focus} = e \times \text{distance to directrix}$

Focus $(a,0)$, directrix $x = -a$: (1 for setup) \sqrt{(x-a)^2 + y^2} = |x + a| \quad (e=1) \tag{2}

Square both sides: (x-a)^2 + y^2 = (x+a)^2 \tag{2} $x^2 - 2ax + a^2 + y^2 = x^2 + 2ax + a^2$ y^2 = 4ax \tag{2}

Eccentricity used: $e = 1$. (1)

Latus rectum: chord through focus perpendicular to axis, $x = a$: $y^2 = 4a^2 \Rightarrow y = \pm 2a$. Endpoints: $(a, 2a)$ and $(a, -2a)$. (2)

Question 2 (12 marks)

(a) Foci $F_1(-c,0)$, $F_2(c,0)$. Definition: \sqrt{(x+c)^2+y^2} + \sqrt{(x-c)^2+y^2} = 2a \tag{1}

Isolate and square: $\sqrt{(x+c)^2+y^2} = 2a - \sqrt{(x-c)^2+y^2}$ (x+c)^2 + y^2 = 4a^2 - 4a\sqrt{(x-c)^2+y^2} + (x-c)^2 + y^2 \tag{2}

Simplify $4cx = 4a^2 - 4a\sqrt{(x-c)^2+y^2}$, so a\sqrt{(x-c)^2+y^2} = a^2 - cx \tag{2}

Square: $a^2[(x-c)^2 + y^2] = a^4 - 2a^2cx + c^2x^2$ $a^2x^2 - 2a^2cx + a^2c^2 + a^2y^2 = a^4 - 2a^2cx + c^2x^2$ (a^2 - c^2)x^2 + a^2 y^2 = a^2(a^2 - c^2) \tag{2}

Let $b^2 = a^2 - c^2$ (valid since $a > c$): dividing by $a^2b^2$: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \qquad b^2 = a^2 - c^2 \tag{1}

(b) Latus rectum: set $x = c$: $\dfrac{c^2}{a^2} + \dfrac{y^2}{b^2} = 1 \Rightarrow y^2 = b^2(1 - c^2/a^2) = b^2\cdot\dfrac{b^2}{a^2}$, so $y = \pm \dfrac{b^2}{a}$. Length $= \dfrac{2b^2}{a}$. (2)

Eccentricity: $e = \dfrac{c}{a} = \dfrac{\sqrt{a^2-b^2}}{a} = \sqrt{1 - \dfrac{b^2}{a^2}}$. (2)

Question 3 (12 marks)

(a) Foci $(\pm c, 0)$, $|r_1 - r_2| = 2a$: \left|\sqrt{(x+c)^2+y^2} - \sqrt{(x-c)^2+y^2}\right| = 2a \tag{1}

Following identical algebra to the ellipse (isolate, square twice): (3) $(c^2 - a^2)x^2 - a^2 y^2 = a^2(c^2 - a^2)$

Here $c > a$, set $b^2 = c^2 - a^2$: (1) \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, \qquad c^2 = a^2 + b^2 \tag{2}

(b) From $\dfrac{y^2}{b^2} = \dfrac{x^2}{a^2} - 1$, so $y = \pm\dfrac{b}{a}\sqrt{x^2 - a^2}$. (1) Asymptotes: $y = \pm \dfrac{b}{a}x$. (2)

Explain out loud: As $|x| \to \infty$, $\sqrt{x^2 - a^2} \approx |x|\sqrt{1 - a^2/x^2} \to |x|$, so the curve approaches $y = \pm\frac{b}{a}x$; the difference between the curve and its asymptote tends to $0$ but never reaches it — the branches hug the lines without touching. (2)

Question 4 (8 marks)

(a) $y^2 = 16x = 4ax \Rightarrow 4a = 16 \Rightarrow a = 4$. Focus $(4, 0)$. (2) Receiver placed at the focus. (1)

(b) Explain out loud: The reflective property states any ray travelling parallel to the axis of a parabola reflects off the surface and passes through the focus (angle of incidence = angle of reflection about the normal). Incoming radio/light waves from a distant source are effectively parallel to the axis; every such ray converges at the focus regardless of where it strikes the dish. Placing the receiver at the focus therefore collects all the reflected energy at a single point, maximising signal strength. (5)

Question 5 (10 marks)

(a) Rule: (2)

• $B^2 - 4AC < 0$: ellipse (circle if $B=0, A=C$)
• $B^2 - 4AC = 0$: parabola
• $B^2 - 4AC > 0$: hyperbola

For $3x^2 - 2xy + 3y^2 - 8=0$: $A=3, B=-2, C=3$. $B^2 - 4AC = 4 - 36 = -32 < 0 \Rightarrow$ ellipse. (2)

(b) (6, 2 for signature, 3 for logic, 1 for returns)

def classify(A, B, C):
    disc = B**2 - 4*A*C
    if disc < 0:
        return "ellipse"
    elif disc == 0:
        return "parabola"
    else:
        return "hyperbola"

Question 6 (8 marks)

(a) Substitute $(ct, c/t)$: $x\cdot y = ct \cdot \dfrac{c}{t} = c^2$ ✓ (1) Rectangular hyperbola has $a = b$, so $e = \sqrt{1 + b^2/a^2} = \sqrt{2}$. (3)

(b) $xy = 9$. Minimise $D^2 = x^2 + y^2 = x^2 + \dfrac{81}{x^2}$. $\dfrac{d}{dx}(D^2) = 2x - \dfrac{162}{x^3} = 0 \Rightarrow x^4 = 81 \Rightarrow x = 3$ (take positive branch). (2) Then $y = 3$, point $(3,3)$; distance $= \sqrt{9+9} = 3\sqrt{2}$. (2)

[
  {"claim":"Ellipse latus rectum with a=5,b=3 equals 2b^2/a = 18/5","code":"a,b=5,3\nresult = (2*b**2/a)==Rational(18,5)"},
  {"claim":"Q5 discriminant for 3x^2-2xy+3y^2 is -32 (<0, ellipse)","code":"A,B,C=3,-2,3\nd=B**2-4*A*C\nresult = (d==-32) and (d<0)"},
  {"claim":"Rectangular hyperbola eccentricity is sqrt(2)","code":"result = sqrt(1+1)==sqrt(2)"},
  {"claim":"Nearest point on xy=9 to origin is (3,3) with distance 3*sqrt(2)","code":"x=symbols('x',positive=True)\nD2=x**2+(9/x)**2\nsol=solve(diff(D2,x),x)\nxv=[s for s in sol if s>0][0]\nresult = (xv==3) and (sqrt(xv**2+(9/xv)**2)==3*sqrt(2))"}
]