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Recognise circles, ellipses, parabolas and hyperbolas from their equations, and find semi-axes, foci, asymptotes and centres.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson four curves that look nothing alike turn out to be one family, separated by the angle at which a plane cuts a cone and, in algebra, by a single sign. You learn to recognise each from its equation, to read the semi-axes out of the denominators, to locate the foci that define an ellipse and a parabola by a distance rule, to find the asymptotes a hyperbola runs alongside, and to complete the square when an equation arrives in a form that hides all of it.
You know the equation of a circle, $(x - h)^2 + (y - k)^2 = r^2$, and you can complete the square to get an equation into that form. You can also read a graph's intercepts off its equation. A conic section is the same kind of object with one number changed — and the change turns a circle into an ellipse, a parabola or a hyperbola.
Conic section: the curve where a plane cuts a double cone — circle, ellipse, parabola or hyperbola.
Semi-axis: half the width ($a$) or half the height ($b$) of an ellipse.
Focus (plural foci): the special point, or pair of points, that defines the curve by a distance rule.
Directrix: the line a parabola is measured against; every point is equidistant from focus and directrix.
Asymptote: a line a hyperbola approaches but never reaches.
Standard form: the equation with the curve centred at the origin and no cross terms.
Slice a double cone and you get a circle, an ellipse, a parabola or a hyperbola, and their standard equations differ by very little. An ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$: two squares added. The semi-axes are $a$ and $b$ — the square roots of the denominators — so $\frac{x^2}{25} + \frac{y^2}{9} = 1$ is $5$ across and $3$ up. Its two foci sit on the longer axis at distance $c$ from the centre with $c^2 = a^2 - b^2$, here $25 - 9 = 16$, so $c = 4$; the defining property is that the two distances to the foci add to the constant $2a$. A hyperbola is the same equation with a minus: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, two branches opening left and right, approaching the asymptotes $y = \pm\frac{b}{a}x$. A parabola has only one squared variable: $x^2 = 4py$, with focus $(0, p)$ and directrix $y = -p$, and every point equidistant from the two. A circle is the ellipse with $a = b$. When an equation arrives untidy — $x^2 + y^2 - 4x + 6y = 12$ — complete the square in each variable to get $(x-2)^2 + (y+3)^2 = 25$ and read it off.
Another way: picture
A double cone with four planes cutting it: horizontally for a circle, gently tilted for an ellipse, parallel to the slanted side for a parabola, and steeply through both halves for a hyperbola — one solid, four curves, distinguished only by the angle of the cut.
Another way: steps
*"The denominators are the semi-axes." They are their squares*. For $\frac{x^2}{25} + \frac{y^2}{9} = 1$ the semi-axes are $5$ and $3$, not $25$ and $9$.
Using $c^2 = a^2 + b^2$ for an ellipse. That is the hyperbola relation. For an ellipse the foci are inside, so $c^2 = a^2 - b^2$ and $c$ is smaller than $a$. Getting these two the wrong way round is the classic slip in this topic.
"The foci are on whichever axis is drawn horizontally." They are always on the longer axis. If the bigger denominator sits under $y^2$, the ellipse is tall and its foci are above and below the centre.
Two squares added, so an ellipse. Semi-axes $\sqrt{25} = 5$ and $\sqrt{9} = 3$.
Roots of the denominators.
The larger is under $x^2$, so it is wide, and the foci are on the $x$-axis: $c^2 = 25 - 9 = 16$, so they are at $(\pm 4, 0)$.
Foci on the longer axis.
Complete the square in $x$: $x^2 - 4x = (x-2)^2 - 4$. In $y$: $y^2 + 6y = (y+3)^2 - 9$.
Half the coefficient, squared.
$(x-2)^2 + (y+3)^2 - 13 = 12$, so $(x-2)^2 + (y+3)^2 = 25$: a circle, centre $(2, -3)$, radius $5$.
The minus sign makes it a hyperbola, opening left and right.
$a = 4$ and $b = 3$, so the asymptotes are $y = \pm\frac{3}{4}x$ and the foci are at $(\pm 5, 0)$, since $c^2 = 16 + 9$.
What curve is $\dfrac{x^2}{64} + \dfrac{y^2}{9} = 1$?
For $\dfrac{x^2}{81} + \dfrac{y^2}{64} = 1$, give the semi-axis along $x$ and then the semi-axis along $y$.
a = a, b = b
For the ellipse $\dfrac{x^2}{3721} + \dfrac{y^2}{3600} = 1$, how far is each focus from the centre?
Each focus is answer from the centre.
The parabola $y = \dfrac{x^2}{24}$ has the form $x^2 = 4py$. Give $p$ and then the $y$-coordinate of the focus.
p = p, focus at (0, f)
The hyperbola $\dfrac{x^2}{16} - \dfrac{y^2}{25} = 1$ has two asymptotes. What is the slope of the one with positive slope?
The positive asymptote has slope answer.
A whispering gallery has an elliptical floor $74$ m long and $24$ m wide. A whisper at one focus is heard clearly at the other. How far from the centre, in metres, is each of those two spots?
Each spot is answer m from the centre.
A satellite dish has a parabolic cross-section $32$ m across the rim and $16$ m deep at the middle. The receiver must sit at the focus. How many metres above the bottom of the dish is that?
Place the receiver answer m above the bottom.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Complete the square to find the centre and radius of $x^2 + y^2 - 30x + 224y = 456$.
centre (h, k), radius r
You can identify any conic from its equation and find its key features. Without looking: in $\frac{x^2}{25} + \frac{y^2}{9} = 1$, what are the semi-axes and where are the foci — and which of $a^2 - b^2$ and $a^2 + b^2$ belongs to an ellipse?
8. Your turn: describe $\frac{x^2}{16} - \frac{y^2}{9} = 1$, step 2