Conic Sections

Q1 (10 marks)

(a) Focus at $(2.5,0)$ ⇒ $a=2.5$. Standard right-opening parabola $y^2=4ax$. $y^2 = 4(2.5)x = 10x.$ (2) (identify $a$ =1, equation =1)

(b) Latus rectum ends: $x=a=2.5$, so $y^2=10(2.5)=25 \Rightarrow y=\pm5$. Rim points: $(2.5, 5)$ and $(2.5, -5)$. (2) Strut length = latus rectum $= 4a = 10$ m. (1)

(c) Point on dish at $x=4$: $y_0^2 = 10(4)=40 \Rightarrow y_0=\sqrt{40}=2\sqrt{10}$. (1) By the reflective property, any ray parallel to the axis reflects through the focus $(2.5,0)$. (1) Distance from line $x=4$ to point $(4, 2\sqrt{10})$: the ray is parallel to axis, so it travels horizontally to the surface point $(4,2\sqrt{10})$ — distance $=4-4=0$ horizontally... take the incoming ray starting on $x=4$: the strike point is $(4,2\sqrt{10})$. Distance surface point → focus: $\sqrt{(4-2.5)^2+(2\sqrt{10}-0)^2}=\sqrt{1.5^2+40}=\sqrt{2.25+40}=\sqrt{42.25}=6.5.$ (2) Total path from line $x=4$ (measured along the incoming horizontal ray from where it crosses $x=4$, which is the strike point itself) to focus $=0+6.5=6.5$ m. (Equivalently, by the focal-distance definition, distance from focus $=x+a=4+2.5=6.5$.) (1)

Q2 (11 marks)

(a) Foci $F_1=(-2,0),\,F_2=(2,0)$, point $P=(2,\tfrac{3\sqrt3}{2})$. $PF_2=\sqrt{(2-2)^2+(\tfrac{3\sqrt3}{2})^2}=\tfrac{3\sqrt3}{2}$. (1) $PF_1=\sqrt{(2+2)^2+(\tfrac{3\sqrt3}{2})^2}=\sqrt{16+\tfrac{27}{4}}=\sqrt{\tfrac{91}{4}}$... let's recompute: $\tfrac{27}{4}=6.75$, $16+6.75=22.75$, $\sqrt{22.75}=4.7697$. Hmm not clean.

Let me use exact: sum $2a = PF_1+PF_2$. (2) $2a = \tfrac{3\sqrt3}{2}+\sqrt{22.75}=2.598+4.770=7.368$.

This is not clean, so the intended clean value: with $c=2$, try $a=4$: then $2a=8$, $b^2=a^2-c^2=12$. Check $P$: $\tfrac{4}{16}+\tfrac{27/4}{12}=0.25+0.5625=0.8125\neq1$. Not on it.

Correct exact sum: $2a=\dfrac{3\sqrt3}{2}+\dfrac{\sqrt{91}}{2}$, hence $a=\dfrac{3\sqrt3+\sqrt{91}}{4}\approx3.684$. (1)

(b) $b^2=a^2-c^2=3.684^2-4=13.57-4=9.57$, $b\approx3.094$. (2) $e=c/a=2/3.684\approx0.543$. (1) Latus rectum $=2b^2/a=2(9.57)/3.684\approx5.20$. (1)

(c) Parametric: $(a\cos\theta,\,b\sin\theta)=(3.684\cos\theta,\,3.094\sin\theta)$. (1) Given point $x=2=3.684\cos\theta\Rightarrow\cos\theta=0.5429\Rightarrow\theta\approx0.9977$ rad. (1) Check $y=3.094\sin(0.9977)=3.094(0.8398)=2.598=\tfrac{3\sqrt3}{2}$ ✓ (so $\theta\approx57.2°$). (1)

Q3 (9 marks)

$A=3,\;B=-2,\;C=3.$ Discriminant $B^2-4AC=4-36=-32<0$ ⇒ ellipse (and since $A=C$, $B\ne0$, it is a tilted ellipse). (3)

Centre: $\partial_x:6x-2y-4=0$, $\partial_y:-2x+6y-4=0$. (2) Solve: from first $3x-y=2$; second $-x+3y=2$. Multiply first by 3: $9x-3y=6$; add second $-x+3y=2$: $8x=8\Rightarrow x=1$, then $y=3x-2=1$. Centre $(1,1)$. (2)

Reality: substitute centre value $F' = F + \tfrac12(Dx_0+Ey_0)= -12+\tfrac12(-4-4)=-12-4=-16<0$ while $A,C>0$; since the quadratic form is positive definite and constant is negative, real points exist ⇒ real (non-degenerate) ellipse. (2)

Q4 (11 marks)

(a) Rectangular hyperbola $xy=16$, so $c^2=16$. Parametric: $(x,y)=(4t,\,4/t),\;t\ne0$. (2) Eccentricity of a rectangular hyperbola $=\sqrt2$ (asymptotes perpendicular ⇒ $a=b$ ⇒ $e=\sqrt{1+b^2/a^2}=\sqrt2$). (1)

(b) Solve $xy=16$ and $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$. From $y=16/x$: $\dfrac{x^2}{25}+\dfrac{256}{9x^2}=1$. (1) Let $u=x^2$: $\dfrac{u}{25}+\dfrac{256}{9u}=1\Rightarrow 9u^2-225u+6400=0$. (1) Discriminant $=225^2-4(9)(6400)=50625-230400=-179775<0$. (1) No real solution ⇒ the curves do not intersect. (2) (The rectangular hyperbola lies entirely outside the ellipse.)

(c) Since there is no intersection point in the first quadrant, the focal-sum verification is vacuous; state clearly no such point exists. (3) (Full marks for correctly justifying non-intersection from (b) and concluding accordingly.)

Ellipse foci: $a=5,b=3,c=\sqrt{25-9}=4$, foci $(\pm4,0)$, $2a=10$ — quoted for completeness.

Q5 (9 marks)

(a) $a^2=9,b^2=16\Rightarrow a=3,b=4$. $c=\sqrt{9+16}=5$. (1) Foci $(\pm5,0)$. (1) $e=c/a=5/3$. (1) Asymptotes: $y=\pm\dfrac{b}{a}x=\pm\dfrac{4}{3}x$. (1)

(b) $x=5$: $\dfrac{25}{9}-\dfrac{y^2}{16}=1\Rightarrow\dfrac{y^2}{16}=\dfrac{16}{9}\Rightarrow y^2=\dfrac{256}{9}\Rightarrow y=\dfrac{16}{3}$ ($y>0$). (2) $P=(5,\tfrac{16}{3})$. Focal radii: to $(5,0)$: $r_1=\sqrt{0+(16/3)^2}=16/3$. To $(-5,0)$: $r_2=\sqrt{100+256/9}=\sqrt{1156/9}=34/3$. (2) Difference $=34/3-16/3=18/3=6=2a$ ✓. (1)

[
  {"claim":"Q1 focal distance to point at x=4 equals 6.5","code":"a=Rational(5,2); x=4; y2=10*x; d=sqrt((x-a)**2+y2); result= simplify(d-Rational(13,2))==0"},
  {"claim":"Q3 discriminant negative (ellipse) and centre (1,1)","code":"A,B,C=3,-2,3; disc=B**2-4*A*C; x,y=symbols('x y'); sol=solve([6*x-2*y-4, -2*x+6*y-4],[x,y]); result=(disc<0) and sol[x]==1 and sol[y]==1"},
  {"claim":"Q4 no real intersection: quadratic discriminant negative","code":"u=symbols('u'); q=9*u**2-225*u+6400; D=225**2-4*9*6400; result=D<0"},
  {"claim":"Q5 difference of focal radii equals 2a=6 at P=(5,16/3)","code":"x=5;y=Rational(16,3); r1=sqrt((x-5)**2+y**2); r2=sqrt((x+5)**2+y**2); result= simplify(r2-r1-6)==0"}
]