Conics

Unit 8 Conics

Key unit competence: Analyse the equations and properties of conic sections by applying them to solve problems.

Key ideas

• A conic is the locus of points $P$ with $\frac{PF}{PD}=e$, where $F$ is the focus, $D$ the distance to the directrix and $e$ the eccentricity: $e=1$ parabola, $0<e<1$ ellipse, $e>1$ hyperbola ($e=0$: circle).
• Parabola $y^2=4ax$: vertex $O$, focus $(a,0)$, directrix $x=-a$, axis $y=0$, latus rectum $4a$. Similarly $x^2=4ay$ has focus $(0,a)$ and directrix $y=-a$.
• Ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ($a>b$): $c^2=a^2-b^2$, foci $(\pm c,0)$, $e=\frac{c}{a}$, directrices $x=\pm \frac{a}{e}$, major axis $2a$, minor axis $2b$. Parametric form $x=a\cos t$, $y=b\sin t$.
• Hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$: $c^2=a^2+b^2$, foci $(\pm c,0)$, $e=\frac{c}{a}>1$, directrices $x=\pm \frac{a}{e}$, asymptotes $y=\pm \frac{b}{a}x$. $x^2-y^2=a^2$ is rectangular (asymptotes $y=\pm x$).
• A general equation $A{x}^2+C{y}^2+Dx+Ey+F=0$ is put in standard form by completing the square, which gives the centre or vertex.
• Polar coordinates: $x=r\cos \theta$, $y=r\sin \theta$, $r^2=x^2+y^2$. A conic with focus at the pole: $r=\frac{\ell }{1+e\cos \theta }$.
• Applications: satellite dishes and headlights (parabolic reflectors send rays through the focus), planetary orbits (ellipses with the Sun at a focus), navigation systems (hyperbolas).

Formulas to know

Common misconceptions

• Using $c^2=a^2+b^2$ for an ellipse (that is the hyperbola).
• For $y^2=8x$, saying the focus is $(8,0)$. Compare with $4a$: $a=2$, focus $(2,0)$.
• For $\frac{x^2}{4}+\frac{y^2}{9}=1$, putting the major axis on the $x$-axis. The larger denominator is under $y^2$.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Find the equation of the parabola with focus $(5,3)$ and directrix $x=1$.

   Show answer

   $PF=PD$: $(x-5{)}^2+(y-3{)}^2=(x-1{)}^2$, so $(y-3{)}^2=8x-24=8(x-3)$. Vertex $(3,3)$.

   Common mistake: Using the distance to the point $(1,0)$ instead of to the line.

2. Identify the conic $r=\frac{4}{2+\cos \theta }$ and give its eccentricity.

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   Divide by 2: $r=\frac{2}{1+\frac{1}{2}\cos \theta }$. $e=\frac{1}{2}<1$: an ellipse.

   Common mistake: Reading $e=1$ without first making the constant term 1.

3. For $9x^2+25y^2=225$, find the axes, eccentricity, foci and directrices.

   Show answer

   $\frac{x^2}{25}+\frac{y^2}{9}=1$: $a=5$, $b=3$, $c=4$; axes 10 and 6; $e=\frac{4}{5}$; foci $(\pm 4,0)$; directrices $x=\pm \frac{25}{4}$.

   Common mistake: Using $c^2=a^2+b^2$.

4. Find the centre, vertices and foci of $4x^2-y^2-8x-4y-4=0$.

   Show answer

   $4(x-1{)}^2-(y+2{)}^2=4-4+4=4$: $\frac{(x-1{)}^2}{1}-\frac{(y+2{)}^2}{4}=1$. Centre $(1,-2)$, $a=1$, $b=2$, $c=\sqrt{5}$. Vertices $(0,-2)$ and $(2,-2)$; foci $(1\pm \sqrt{5},-2)$.

   Common mistake: Sign errors when completing the square for $-y^2-4y$.

5. Find the equation of the ellipse with centre $O$, a focus at $(0,4)$ and $e=\frac{4}{5}$.

   Show answer

   Major axis vertical, $c=4$, $a=\frac{c}{e}=5$, $b^2=25-16=9$: $\frac{x^2}{9}+\frac{y^2}{25}=1$.

   Common mistake: Putting 25 under $x^2$.

6. Show that the line $y=x+1$ is tangent to the parabola $y^2=4x$.

   Show answer

   $(x+1{)}^2=4x$ gives $x^2-2x+1=0=(x-1{)}^2$: one double root, so the line touches the parabola at $(1,2)$.

   Common mistake: Finding the intersection but not checking that it is a double root.

7. Convert $x^2+y^2-6y=0$ to polar form.

   Show answer

   $r^2-6r\sin \theta =0$, so $r=6\sin \theta$ (a circle with centre $(0,3)$, radius 3).

   Common mistake: Replacing $y$ by $r\cos \theta$.

8. A parabolic satellite dish is 2 m wide and 0.25 m deep. How far from the vertex should the receiver be placed?

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   Model $x^2=4ay$ through $(1,0.25)$: $1=4a\times 0.25$, $a=1$. The receiver (focus) is 1 m from the vertex.

   Common mistake: Using the full width 2 instead of the half-width 1.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Find the asymptotes of $\frac{x^2}{9}-\frac{y^2}{16}=1$.	a. $(\pm 5,0)$ and $\frac{5}{3}$
   2. Find the tangent to $\frac{x^2}{4}+\frac{y^2}{3}=1$ at $\left(1,\frac{3}{2}\right)$.	b. $\frac{x}{4}+\frac{y}{2}=1$
   3. Find the length of the latus rectum of $y^2=12x$.	c. $y=\pm \frac{4}{3}x$
   4. Find the foci and eccentricity of $\frac{x^2}{9}-\frac{y^2}{16}=1$.	d. $12$

   Show answer

   1 → c ($y=\pm \frac{4}{3}x$) · 2 → b ($\frac{x}{4}+\frac{y}{2}=1$) · 3 → d ($12$) · 4 → a ($(\pm 5,0)$ and $\frac{5}{3}$)

   • Step 1. For $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ the asymptotes are $y=\pm \frac{b}{a}x$.
     Step 2. $a=3$, $b=4$.
   • Step 1. Tangent at $(x_0,y_0)$: $\frac{x{x}_0}{a^2}+\frac{y{y}_0}{b^2}=1$.
     Step 2. $\frac{x}{4}+\frac{\frac{3}{2}y}{3}=\frac{x}{4}+\frac{y}{2}=1$.
   • Step 1. $4a=12$, so $a=3$; at $x=3$, $y^2=36$, $y=\pm 6$.
     Step 2. Length $=6-(-6)=12$.
   • Step 1. Hyperbola: $c^2=a^2+b^2=9+16=25$, $c=5$.
     Step 2. $e=\frac{c}{a}=\frac{5}{3}$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   vertex $(2,2)$, focus $(4,2)$  ·  $\frac{x^2}{25}+\frac{y^2}{16}=1$  ·  $\frac{x^2}{16}+\frac{y^2}{25}=1$  ·  $\frac{x^2}{4}-\frac{y^2}{5}=1$
   1. The ellipse with foci $(\pm 3,0)$ and major axis 10 is .
   2. The hyperbola with vertices $(\pm 2,0)$ and foci $(\pm 3,0)$ is .
   3. For $y^2-4y-8x+20=0$, the vertex and focus are .
   Show answer

   1: $\frac{x^2}{25}+\frac{y^2}{16}=1$ · 2: $\frac{x^2}{4}-\frac{y^2}{5}=1$ · 3: vertex $(2,2)$, focus $(4,2)$

   • Step 1. $2a=10\Rightarrow a=5$; $c=3$.
     Step 2. $b^2=a^2-c^2=25-9=16$: $\frac{x^2}{25}+\frac{y^2}{16}=1$.
   • Step 1. $a=2$, $c=3$.
     Step 2. $b^2=c^2-a^2=9-4=5$: $\frac{x^2}{4}-\frac{y^2}{5}=1$.
   • Step 1. $(y-2{)}^2=8x-16=8(x-2)$: vertex $(2,2)$, $4a=8$, $a=2$.
     Step 2. It opens to the right: focus $(2+2,2)=(4,2)$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Find the foci of $\frac{x^2}{16}+\frac{y^2}{9}=1$.	a. $\sqrt{2}$
   2. Find the eccentricity of $\frac{x^2}{16}+\frac{y^2}{9}=1$.	b. $\frac{x^2}{9}+\frac{y^2}{4}=1$
   3. Eliminate $t$ from $x=3\cos t$, $y=2\sin t$.	c. $(\pm \sqrt{7},0)$
   4. Find the eccentricity of the rectangular hyperbola $xy=8$.	d. $\frac{\sqrt{7}}{4}$

   Show answer

   1 → c ($(\pm \sqrt{7},0)$) · 2 → d ($\frac{\sqrt{7}}{4}$) · 3 → b ($\frac{x^2}{9}+\frac{y^2}{4}=1$) · 4 → a ($\sqrt{2}$)

   • Step 1. Ellipse: $c^2=a^2-b^2=16-9=7$.
     Step 2. The major axis is along $x$, so the foci are $(\pm \sqrt{7},0)$.
   • Step 1. $c=\sqrt{7}$ and $a=4$.
     Step 2. $e=\frac{c}{a}=\frac{\sqrt{7}}{4}$.
   • Step 1. $\cos t=\frac{x}{3}$ and $\sin t=\frac{y}{2}$.
     Step 2. ${\cos }^{2}t+{\sin }^{2}t=1$ gives $\frac{x^2}{9}+\frac{y^2}{4}=1$.
   • Step 1. A rectangular hyperbola has $a=b$.
     Step 2. $e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{2}$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $\frac{\sqrt{6}}{3}$  ·  a hyperbola ($e=2$)  ·  a circle  ·  $(2,2)$
   1. The conic $r=\frac{6}{1+2\cos \theta }$ is .
   2. The centre of the ellipse $x^2+3y^2-4x-12y+13=0$ is .
   3. The eccentricity of the ellipse $x^2+3y^2-4x-12y+13=0$ is .
   Show answer

   1: a hyperbola ($e=2$) · 2: $(2,2)$ · 3: $\frac{\sqrt{6}}{3}$

   • Step 1. Compare with $r=\frac{\ell }{1+e\cos \theta }$: $e=2$.
     Step 2. $e>1$ means a hyperbola.
   • Step 1. $(x^2-4x)+3(y^2-4y)=-13$.
     Step 2. $(x-2{)}^2+3(y-2{)}^2=-13+4+12=3$: centre $(2,2)$.
   • Step 1. $\frac{(x-2{)}^2}{3}+(y-2{)}^2=1$: $a^2=3$, $b^2=1$, $c^2=2$.
     Step 2. $e=\frac{c}{a}=\sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3}$.

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