Why are ellipses, parabolas and hyperbolas called conic curves?

Because each one is the shape of a cross-section made by cutting a cone with a flat plane. The angle at which the plane cuts decides which curve appears, and the eccentricity is the single number that records which one you have.

Define an asymptote of a hyperbola.

An asymptote is a straight line that the curve gets arbitrarily close to as it runs away to infinity, without ever reaching it. A hyperbola has two of them, crossing at the origin, and each branch is trapped in the wedge between them.

A LevelFurther MathsEdexcelFurther Pure 1FlashcardsCoordinate SystemsProperties of Ellipses, Parabolas & Hyperbolas

Properties of Ellipses, Parabolas & Hyperbolas (Edexcel A Level Further Maths: Further Pure 1): Flashcards

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FrontProperties of Ellipses
Why are ellipses, parabolas and hyperbolas called conic curves?

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Why are ellipses, parabolas and hyperbolas called conic curves?
Because each one is the shape of a cross-section made by cutting a cone with a flat plane.
The angle at which the plane cuts decides which curve appears, and the eccentricity is the single number that records which one you have.
Define the eccentricity of an ellipse.
The eccentricity $e$ measures how stretched an ellipse is, and satisfies $0 \leq e < 1$ .
At $e = 0$ the ellipse is a perfect circle, and as $e$ gets closer to $1$ the ellipse becomes flatter and flatter.
What is the difference between an ellipse's major axis and its semi-major axis?
The major axis is the longer of the two axes, running right across the ellipse, of length $2 a$ when $a > b$ .
The semi-major axis is half of it, running from the centre out to the edge, of length $a$ .
On the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ , what is the difference between the point $(3 \cos θ, 2 \sin θ)$ and the point $(3, 0)$ ?
Both lie on the ellipse, but $(3 \cos θ, 2 \sin θ)$ is a general point, which moves round the curve as $θ$ varies.
$(3, 0)$ is a fixed point, one particular position, and is what the general point becomes when $θ = 0$ .
$P$ is any point on an ellipse, $F$ is a focus, and $D$ is the nearest point to $P$ on the matching directrix. What does the focus-directrix property say about $P F$ and $P D$ ?
Their ratio is always the eccentricity, $\frac{P F}{P D} = e$ , wherever $P$ sits on the curve.
It is often used rearranged as $P F = e P D$ , and the same value of $e$ comes out when the other focus and its own directrix are used instead.
True or False?
Every point on an ellipse is the same total distance from the two foci.
True.
For every point $P$ on the ellipse, $P F + P F'$ is equal to the length of the major axis, which is $2 a$ in the case $a > b$ .
It follows from applying the focus-directrix property at each focus in turn and adding, because the two directrices are a fixed distance apart.
An ellipse has $a < b$ , so its major axis is vertical. Fill in its directrices and its foci:
$directrices y = \pm______and foci (0, \pm______)$
The completed line is:
$directrices y = \pm \frac{b}{e} and foci (0, \pm b e)$
Everything swaps from horizontal to vertical and $a$ swaps with $b$ , so the eccentricity now comes from $a^{2} = b^{2} (1 - e^{2})$ .
A parabola has eccentricity $e = 1$ . What does that say about the distance from a point on the parabola to the focus and to the directrix?
They are equal, so $P F = P D$ for every point $P$ on the curve.
A parabola is therefore exactly the set of points that are equidistant from a fixed point, the focus, and a fixed line, the directrix.
Define the directrix of a parabola.
The directrix is the fixed straight line that, together with the focus, defines the curve; for $y^{2} = 4 a x$ it is the vertical line $x = - a$ .
It sits on the opposite side of the vertex from the focus, the same distance $a$ away.
What are the vertex and the line of symmetry of the parabola $y^{2} = 4 a x$ ?
The vertex is at the origin $(0, 0)$ and the line of symmetry is $y = 0$ , the $x$ -axis.
The curve looks like $y = x^{2}$ turned through $90 °$ clockwise, so it opens to the right rather than upwards.
Fill in the two gaps in this elimination of the parameter from $x = a t^{2}$ and $y = 2 a t$ :
$t = \frac{y}{______} so x = a \times \frac{y^{2}}{4 a^{2}} which rearranges to y^{2} =______x$
The completed working is:
$t = \frac{y}{2 a} so x = a \times \frac{y^{2}}{4 a^{2}} which rearranges to y^{2} = 4 a x$
No identity is needed anywhere here, because the parameter can be made the subject of the linear equation directly.
The general point on the parabola $y^{2} = 4 a x$ is $(a t^{2}, 2 a t)$ . Which values of $t$ are allowed, and what does each one give?
Every real value of $t$ is allowed, with nothing excluded, and each one gives exactly one point of the parabola.
Unlike the trigonometric parametrisations of the other conics, the parameter here is not an angle, so there is no restricted range to remember.
True or False?
A parabola that is drawn narrower has a smaller eccentricity than one drawn wider.
False.
Every parabola has $e = 1$ exactly, whatever the value of $a$ .
Changing $a$ moves the focus and the directrix further apart and scales the whole curve up, but it does not change the shape, so the eccentricity is untouched.
A point $P (x, y)$ on a parabola is equidistant from the focus $(a, 0)$ and the directrix $x = - a$ . How does that give the Cartesian equation $y^{2} = 4 a x$ ?
Write each distance out: $P F$ comes from Pythagoras as $\sqrt{{(x - a)}^{2} + y^{2}}$ , while $P D$ is simply the horizontal distance $x + a$ .
Setting the two squares equal, the $x^{2}$ and $a^{2}$ terms cancel from both sides and what is left is $y^{2} = 4 a x$ .
Which $x$ -values does each branch of $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ occupy?
The positive- $x$ branch has $x \geq a$ and the negative- $x$ branch has $x \leq - a$ .
Nothing lies in the gap between them, because rearranging to $\frac{x^{2}}{a^{2}} = 1 + \frac{y^{2}}{b^{2}} \geq 1$ forces $| x | \geq a$ .
Define an asymptote of a hyperbola.
An asymptote is a straight line that the curve gets arbitrarily close to as it runs away to infinity, without ever reaching it.
A hyperbola has two of them, crossing at the origin, and each branch is trapped in the wedge between them.
True or False?
The two asymptotes of a hyperbola always meet at right angles.
False.
They are $y = \frac{b}{a} x$ and $y = - \frac{b}{a} x$ , whose gradients multiply to give $- \frac{b^{2}}{a^{2}}$ .
That is equal to $- 1$ only when $a = b$ , so for most hyperbolas the asymptotes are not perpendicular at all.
Where do the asymptotes $y = \pm \frac{b}{a} x$ of a hyperbola come from?
Rearranging the equation of the curve gives $y = \pm \frac{b x}{a} \sqrt{1 - \frac{a^{2}}{x^{2}}}$ .
As $x$ grows the term $\frac{a^{2}}{x^{2}}$ tends to zero, so the square root tends to $1$ and the curve settles onto the straight lines $y = \pm \frac{b}{a} x$ .
For a hyperbola, where do the foci and the directrices sit relative to the two branches?
Each focus $(\pm a e, 0)$ lies inside the hollow of its own branch, because $e > 1$ makes $a e$ bigger than $a$ .
Both directrices $x = \pm \frac{a}{e}$ lie in the gap between the branches, because the same condition makes $\frac{a}{e}$ smaller than $a$ .
For the hyperbola parametrised as $(a \sec θ, b \tan θ)$ , fill in which branch each range of $θ$ describes:
$- \frac{π}{2} < θ < \frac{π}{2} gives the______- x branch, and \frac{π}{2} < θ \leq π gives the______- x branch$
The completed line is:
$- \frac{π}{2} < θ < \frac{π}{2} gives the positive- x branch, and \frac{π}{2} < θ \leq π gives the negative- x branch$
The branch follows the sign of $\sec θ$ , which is positive over the first range and negative over the second.
Both $θ = \frac{π}{2}$ and $θ = - \frac{π}{2}$ have to be excluded, since $\sec θ$ and $\tan θ$ are undefined there.
A hyperbola has parametric coordinates $(\pm a \cosh θ, b \sinh θ)$ . Why is the $\pm$ needed on the $x$ -coordinate?
Because $\cosh θ$ is never less than $1$ , so $a \cosh θ$ is always positive and on its own would reach only the positive- $x$ branch.
The minus sign is what supplies the negative- $x$ branch, and each choice of sign then traces one whole branch.
Why does a hyperbola have two different sets of parametric equations, where the other conics have only one?
Because the equation $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ needs an identity of the shape 'square minus square equals one', and there are two of those available, one hyperbolic and one trigonometric.
Either one reduces the equation to $1 = 1$ , so both describe the same hyperbola and either may be used.
Why is the curve $x y = c^{2}$ called a rectangular hyperbola?
Because its two asymptotes, which are the $x$ -axis and the $y$ -axis, are perpendicular to one another.
A general hyperbola's asymptotes cross at some other angle, so the right angle is exactly what makes this one a special case.
True or False?
The graph of $y = \frac{1}{x}$ is a rectangular hyperbola.
True.
Rearranged it reads $x y = 1$ , which is $x y = c^{2}$ with $c = 1$ .
The familiar reciprocal graph is therefore a conic, and not a curve of some different kind at all.
What are the lines of symmetry of the rectangular hyperbola $x y = c^{2}$ ?
They are $y = x$ and $y = - x$ .
Swapping $x$ and $y$ leaves $x y = c^{2}$ unchanged, which is reflection in $y = x$ , and replacing $(x, y)$ by $(- y, - x)$ leaves it unchanged too, which is reflection in $y = - x$ .
The rectangular hyperbola $x y = c^{2}$ has general point $(c t, \frac{c}{t})$ . Fill in the value $t$ can never take, and what would go wrong:
$t \neq______because \frac{c}{t} would then be______for that point$
The completed line is:
$t \neq 0 because \frac{c}{t} would then be undefined for that point$
So the parameter runs over every real number except zero, which fits the curve never meeting either axis.
For a rectangular hyperbola $x y = c^{2}$ , the foci lie on the line $y = x$ rather than on a coordinate axis. Why?
Because the curve's axis of symmetry is $y = x$ , not the $x$ -axis: the two branches sit in the first and third quadrants, facing each other along that line.
The foci always lie on a conic's axis of symmetry, so here they are at $(\pm \sqrt{2} c, \pm \sqrt{2} c)$ , and the directrices $x + y = \pm \sqrt{2} c$ run perpendicular to it.
A rectangular hyperbola has equation $x y = 36$ . What is $c$ , and what are the coordinates of its foci?
Comparing with $x y = c^{2}$ gives $c = 6$ , taking the positive square root because $c > 0$ .
The foci are then $(6 \sqrt{2}, 6 \sqrt{2})$ and $(- 6 \sqrt{2}, - 6 \sqrt{2})$ .

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