Slice a Cone at Different Angles and Four Different Curves Appear

Take two identical cones joined tip to tip along a vertical axis — a double-napped cone. Slice it with a flat plane, and the edge of the cut is a curve. Depending on the angle of the slice, that curve is a circle, an ellipse, a parabola or a hyperbola — so these are called conic sections.

This part covers how the angle decides the curve, the degenerate cases, the equation of a circle, and circles from given conditions.

**If the cone's generating line makes a semi-vertical angle $\alpha$ with the axis and the cutting plane makes an angle $\beta$ with the axis, then $\beta =90^{\circ }$ gives a circle, $\alpha <\beta <90^{\circ }$ an ellipse, $\beta =\alpha$ a parabola, and $0^{\circ }\le \beta <\alpha$ a hyperbola.

- Circle — plane perpendicular to the axis
- Ellipse — plane tilted, but steeper than a generator
- Parabola — plane parallel to a generator
- Hyperbola — plane steep enough to cut both nappes

Worked example.** For a cone with $\alpha =30^{\circ }$, classify planes at $\beta =90^{\circ },60^{\circ },30^{\circ },15^{\circ }$.

$$90^{\circ }:\text{circle}30^{\circ }<60^{\circ }<90^{\circ }:\text{ellipse}30^{\circ }=\alpha :\text{parabola}15^{\circ }<30^{\circ }:\text{hyperbola}$$

An everyday example. A torch shone straight at a wall makes a circle of light, and tilting it stretches the patch into other conics.

The substance. A hyperbola has two separate branches because the plane cuts both halves of the double cone.

When the cutting plane passes through the vertex of the cone, the section shrinks to a point, a single straight line, or a pair of intersecting lines — the degenerate conics.

- **$\alpha <\beta \le 90^{\circ }$ through the vertex — a point, the degenerate circle or ellipse
- $\beta =\alpha$ through the vertex — a straight line, one generator, the degenerate parabola
- $0^{\circ }\le \beta <\alpha$ through the vertex — a pair of intersecting lines, the degenerate hyperbola

Worked examples in equations.**

$$x^2+y^2=0\text{is only the point }(0,0)$$
$$x^2-y^2=0\Rightarrow (x-y)(x+y)=0\text{the pair of lines }y=x\text{ and }y=-x$$

A circle $(x-h{)}^2+(y-k{)}^2=r^2$ with $r=0$ also shrinks to the single point $(h,k)$.

An everyday example. An ice-cream cone cut with a knife exactly through its tip shows two straight edges meeting at a point — a pair of intersecting lines.

The link. Degenerate conics are limiting cases of the ordinary ones.

**A point $(x,y)$ at distance $r$ from the centre $(h,k)$ satisfies $(x-h{)}^2+(y-k{)}^2=r^2$; for $x^2+y^2+2gx+2fy+c=0$, the centre is $(-g,-f)$ and the radius is $\sqrt{g^2+f^2-c}$.

Worked example 1 — writing the equation.** Centre $(-2,3)$, radius $4$:

$$(x+2{)}^2+(y-3{)}^2=16x^2+y^2+4x-6y-3=0$$

Worked example 2 — completing the square. $x^2+y^2-8x+10y-12=0$:

$$(x-4{)}^2+(y+5{)}^2=12+16+25=53$$

Centre $(4,-5)$, radius $\sqrt{53}$.

Worked example 3 — unequal leading coefficient. $2x^2+2y^2-x=0$. Divide by $2$:

$$x^2+y^2-\frac{x}{2}=0{\left(x-\frac{1}{4}\right)}^2+y^2=\frac{1}{16}$$

Centre $\left(\frac{1}{4},0\right)$, radius $\frac{1}{4}$.

An everyday example. A mobile tower's coverage boundary is a circle centred on the tower.

The substance. **An equation represents a circle only if $x^2$ and $y^2$ have equal coefficients, there is no $xy$ term, and $g^2+f^2-c\ge 0$.**

**With a centre and radius use the standard form; with the ends of a diameter use $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$; with points on the circle, substitute them into the general equation and solve.

Worked example 1 — diameter.** Ends $(2,-3)$ and $(-2,5)$:

$$(x-2)(x+2)+(y+3)(y-5)=0x^2+y^2-2y-19=0$$

Check: centre $(0,1)$ and radius $\sqrt{4+16}=\sqrt{20}$ give the same equation.

Worked example 2 — a right angle. Through $(0,0)$, $(6,0)$ and $(0,8)$. The angle at the origin is $90^{\circ }$, so the segment from $(6,0)$ to $(0,8)$ is a diameter:

$$(x-6)x+y(y-8)=0x^2+y^2-6x-8y=0$$

Worked example 3 — centre on a line. Through $(2,-2)$ and $(3,4)$ with centre on $x+y=2$. Let the centre be $(h,2-h)$; equal distances give

$$(2-h{)}^2+(h-4{)}^2=(3-h{)}^2+(h+2{)}^{2}20-12h=13-2hh=0.7$$

Centre $(0.7,1.3)$ and $r^2=1.69+10.89=12.58$, so $x^2+y^2-1.4x-2.6y-10.4=0$.

An everyday example. A circular park path through three lamp posts is found this way.

The substance. Three non-collinear points fix exactly one circle.