Slicing a cone
Tonight's idea
Cut a double cone (two cones tip to tip) with a flat plane. Cut straight across and you get a circle. Tilt the plane and the circle stretches into an ellipse. Tilt it parallel to the slanted side and the curve stops closing up: a parabola. Tilt further and the plane hits both halves: a hyperbola, two branches.
There is one definition that covers all three. Fix a point \(F\) (the focus) and a line \(d\) (the directrix), and take all points \(P\) with
\(\dfrac{\text{distance from } P \text{ to } F}{\text{distance from } P \text{ to the line } d} = e\), the eccentricity. If \(e \lt 1\) you get an ellipse, \(e = 1\) a parabola, \(e \gt 1\) a hyperbola.
An ellipse also comes from two pins and a loop of string: the sum of the distances to the pins never changes. Call that sum \(2a\) and the pin separation \(2c\). The longest radius is \(a\). The top point is \(a\) from each focus (its two distances are equal and add to \(2a\)), so the shortest radius \(b\) is a leg of a right triangle with hypotenuse \(a\) and other leg \(c\):
\(a^2 = b^2 + c^2\), \(e = c/a\), and the standard ellipse is \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\) with foci \((\pm c, 0)\).
Sound leaving one focus of an ellipse bounces to the other focus: a whispering gallery. A parabola sends everything arriving parallel to its axis to the focus: a satellite dish. Demo: mathdetours.com/demos/conics.html.
Watch one: where does \(y = x^2\) come from?
Focus \(F = (0, p)\), directrix \(y = -p\). A point \((x, y)\) is on the parabola when its distances to \(F\) and to the line are equal. Distance to the line is \(y + p\); to the focus, \(\sqrt{x^2 + (y - p)^2}\). Square both:
\[x^2 + (y - p)^2 = (y + p)^2.\]
The \(y^2\) and \(p^2\) cancel, leaving \(x^2 = 4py\), that is \(y = \dfrac{x^2}{4p}\). So \(y = x^2\) has \(4p = 1\): focus \((0, \tfrac14)\), directrix \(y = -\tfrac14\).
Your turn
- Find the focus and directrix of \(y = x^2/12\). Then write the equation of the parabola with focus \((0, 3)\) and directrix \(y = -3\).
- For the ellipse \(\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1\): write down \(a\) and \(b\), find \(c\), the two foci and the eccentricity. Check that \((0, 3)\) really is a total distance \(2a\) from the two foci.
- A satellite dish is 1 m across and 10 cm deep at the centre. Model its cross-section as \(y = x^2/(4p)\) with the vertex at the origin. How far above the vertex should the receiver sit? (Which rim point do you know the coordinates of?)
- A whispering gallery is an elliptical room 40 m long and 24 m wide. Where should two people stand to hear each other whisper? Give the positions measured from the centre of the room along its long axis. How far apart are they?
Stretch: how oval is Earth's orbit?
Earth's orbit has eccentricity \(0.0167\). If you drew it as an ellipse with \(a = 10\) cm, how far from the centre would the Sun sit, and how much shorter than \(a\) is \(b\)? Is the orbit visibly oval?
Answers
- \(4p = 12\) so \(p = 3\): focus \((0, 3)\), directrix \(y = -3\). The second parabola is the same one, \(y = x^2/12\).
- \(a = 5\), \(b = 3\), \(c = \sqrt{25 - 9} = 4\). Foci \((\pm 4, 0)\), \(e = 0.8\). From \((0, 3)\) each focus is \(\sqrt{16 + 9} = 5\) away, total \(10 = 2a\).
- The rim passes through \((0.5, 0.1)\): \(0.1 = 0.25/(4p)\) gives \(4p = 2.5\), \(p = 0.625\) m. The receiver sits \(62.5\) cm above the vertex, well above the rim, which is why it sticks out on an arm.
- \(a = 20\), \(b = 12\), \(c = \sqrt{400 - 144} = 16\). Stand at the foci, \(16\) m either side of the centre along the long axis: \(32\) m apart.
Stretch. \(c = ea = 0.167\) cm, so the Sun sits under 2 mm from the centre. \(b = a\sqrt{1 - e^2} = 9.9986\) cm, shorter by about \(0.0014\) cm, a hundredth of a millimetre. Drawn to scale the orbit is a circle to the eye; what is noticeably off-centre is the Sun.