Standard Conic Forms and Features

Overview

This branch is the feature-reading reference for Conic Sections. Use it when the question gives a conic in recognisable standard form, or after completing the square has revealed that standard form.

The central habit is to read the equation as geometry: centre or vertex, symmetry, axes, vertices, radius, semi-axis lengths, and asymptotes where relevant.

Enrichment: Why These Curves Are Called Conics

The word conic comes from cone.

A conic section is a curve formed when a flat plane cuts a right circular double cone. The double cone has two halves, called nappes. Depending on how the plane cuts the cone, a different curve is formed.

This gives a single geometric origin for four curves that may otherwise seem unrelated:

• a circle is formed when the plane cuts the cone horizontally
• an ellipse is formed when the plane cuts through one nappe at a slant, without being parallel to the cone surface
• a parabola is formed when the plane cuts through one nappe parallel to the cone surface
• a hyperbola is formed when the plane cuts through both nappes

The important idea is not just the picture of the cone. The important idea is that these four curves belong to one family.

This matters because, without the cone picture, the topic can feel like four separate lists of equations:

$$\text{circle},\text{ellipse},\text{parabola},\text{hyperbola}.$$

But the cone picture tells us that they are connected. They are different outcomes of the same geometric process.

For H2 graphing, we usually do not work directly with the three-dimensional cone. Instead, we work with equations in the coordinate plane. The goal is therefore to connect two views of the same object:

• the geometric view: what kind of curve it is and what features determine its shape
• the algebraic view: how those features appear in the equation

So the cone explains why these curves form one family, while the standard equations show how to study them precisely.

What to Look For When Sketching a Conic

Sketching a conic is not mainly about plotting many points.

A good sketch comes from identifying the few structural features that determine the curve. Once these features are known, the shape is mostly forced.

Different conics require different features.

For a circle, identify:

• the centre
• the radius
• axial intercepts, if useful

For example, from

$$(x-h{)}^2+(y-k{)}^2=r^2,$$

the centre is $(h,k)$ and the radius is $r$.

For an ellipse, identify:

• the centre
• the horizontal and vertical semi-axis lengths
• axial intercepts, if useful

For example, from

$$\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1,$$

the centre is $(h,k)$, the horizontal semi-axis length is $a$, and the vertical semi-axis length is $b$.

For a hyperbola, identify:

• the centre
• the vertices
• the direction of the branches
• the asymptotes
• axial intercepts, if useful

For a hyperbola, the asymptotes are not optional decoration. They control the long-range direction of the branches and should be drawn clearly, usually as dashed lines.

For a parabola, identify:

• the vertex
• the line of symmetry
• the direction of opening
• axial intercepts, if useful

For example,

$$(y-k{)}^2=4a(x-h)$$

has vertex $(h,k)$ and opens horizontally, while

$$(x-h{)}^2=4a(y-k)$$

has vertex $(h,k)$ and opens vertically.

The equation of the conic should usually be shown beside the sketch. This reminds us that the drawing is not just a rough picture. It is a visual interpretation of a precise algebraic equation.

Intercepts should be checked, not guessed. Some conics have two intercepts, some have one, some have none, and some just touch an axis. These details should come from the equation, not from the appearance of a rough sketch alone.

A reliable sketching workflow is:

1. Rewrite the equation into standard form if necessary.
2. Identify the type of conic.
3. Read the centre or vertex from the standard form.
4. Mark the structural features that determine the curve.
5. Check axial intercepts if they are needed.
6. Draw the curve using these features.

The central habit is:

Do not sketch first and interpret later.
Read the equation first, then let the geometry guide the sketch.

Circle

A circle is the set of all points in a plane that are the same distance from a fixed point.

The fixed point is called the centre, and the fixed distance is called the radius.

This definition is important because it explains the equation of a circle. If the centre is

$$C(h,k)$$

and a point on the circle is

$$P(x,y),$$

then the condition for $P$ to lie on the circle is

$$CP=r.$$

Using the distance formula,

$$\sqrt{(x-h{)}^2+(y-k{)}^2}=r.$$

Squaring both sides gives the standard form of a circle:

$$(x-h{)}^2+(y-k{)}^2=r^2.$$

So the equation is not an arbitrary formula. It is just the distance condition written algebraically.

The simplest case occurs when the centre is the origin. If $(h,k)=(0,0)$, then the equation becomes

$$x^2+y^2=r^2.$$

The general form

$$(x-h{)}^2+(y-k{)}^2=r^2$$

is the same circle shifted so that its centre is now $(h,k)$ instead of $(0,0)$.

This standard form should be read geometrically:

• $(h,k)$ is the centre of the circle
• $r$ is the radius
• $r^2$ appears on the right-hand side, so the radius is the square root of that value

For example, from

$$(x-2{)}^2+(y+1{)}^2=9,$$

we read

$$h=2,k=-1,r=3.$$

So the centre is

$$(2,-1)$$

and the radius is $3$.

A circle has uniform symmetry around its centre. In particular, it is symmetric about every line passing through the centre. For coordinate sketches, the most useful symmetry lines are often

$$x=h$$

and

$$y=k.$$

These are the vertical and horizontal lines through the centre.

When sketching a circle, first mark the centre. Then use the radius to locate the four extreme points:

$$(h+r,k),(h-r,k),(h,k+r),(h,k-r).$$

For

$$(x-2{)}^2+(y+1{)}^2=9,$$

the four extreme points are

$$(5,-1),(-1,-1),(2,2),(2,-4).$$

These points help determine the shape without plotting many random points.

A circle may not be given in standard form at first. For example,

$$x^2+y^2-4x+2y=4$$

does not immediately show the centre or radius. Completing the square gives

$$(x-2{)}^2+(y+1{)}^2=9.$$

Now the geometry is visible: the centre is $(2,-1)$ and the radius is $3$.

This is why completing the square is a structural step, not just algebraic tidying. It reveals the hidden centre and radius.

In a general quadratic equation, equal same-sign coefficients of $x^2$ and $y^2$ suggest a circle. For example, both $x^2$ and $y^2$ have coefficient $1$ in

$$x^2+y^2-4x+2y=4.$$

However, this is only a clue. The equation should still be rewritten into standard form to check that the radius is real.

For instance, if completing the square leads to

$$(x-h{)}^2+(y-k{)}^2=-1,$$

then there is no real circle, because the left-hand side is always non-negative.

So for a circle, the main habit is:

Read the equation as a constant-distance condition from a centre.

Caption: In circle form, the centre and radius are read directly from the completed-square equation.

What the algebra tells you

For a circle, the squared terms have a very special pattern.

In an equation of the form

$$A{x}^2+B{y}^2+Cx+Dy+E=0,$$

a circle is possible only when the coefficients of $x^2$ and $y^2$ are equal and have the same sign.

For example,

$$x^2+y^2-4x+6y-12=0$$

has equal coefficients for $x^2$ and $y^2$, so it may represent a circle.

But this coefficient pattern is only a clue. It does not yet tell us the centre or radius. To see the actual circle, we must complete the square:

$$x^2-4x+y^2+6y=12,$$ $$(x-2{)}^2+(y+3{)}^2=25.$$

Now the geometry is visible. The centre is $(2,-3)$ and the radius is $5$.

There is also an important warning. Equal same-sign coefficients suggest a circle, but the final equation must still have a non-negative value on the right-hand side. For instance,

$$(x-2{)}^2+(y+3{)}^2=-1$$

does not represent a real circle, because the left-hand side cannot be negative.

So the algebra gives a two-step reading:

1. equal same-sign coefficients suggest a circle
2. completing the square confirms the circle and reveals its centre and radius

What to identify

When sketching or interpreting a circle, the main features are:

• the centre
• the radius
• the axial intercepts, if they exist
• whether the circle passes through the origin
• whether the circle just touches an axis

The centre and radius are the essential features. They determine the basic position and size of the circle.

The intercepts tell us where the circle meets the coordinate axes. They should be checked from the equation, not guessed from a rough sketch.

It is also useful to know whether the circle passes through the origin. For a circle with centre $(h,k)$ and radius $r$,

$$(x-h{)}^2+(y-k{)}^2=r^2,$$

the origin lies on the circle exactly when

$$h^2+k^2=r^2.$$

This is because substituting $(0,0)$ into the equation gives

$$(0-h{)}^2+(0-k{)}^2=r^2.$$

Similarly, the circle just touches the $y$-axis when the horizontal distance from the centre to the $y$-axis equals the radius:

$$|h|=r.$$

It just touches the $x$-axis when the vertical distance from the centre to the $x$-axis equals the radius:

$$|k|=r.$$

These checks are not just extra details. They help locate the circle relative to the coordinate axes.

The useful habit is:

Do not only ask what type of conic it is.
Ask where the curve sits in the coordinate plane.

Ellipse

A circle has the same radius in every direction from its centre. An ellipse is similar in spirit, but it is stretched differently in different directions.

So an ellipse can be thought of as a circle-like curve with a horizontal size and a vertical size.

Geometrically, an ellipse is the set of all points whose sum of distances from two fixed points is constant. These two fixed points are called the foci.

If $P(x,y)$ is a point on the ellipse and the foci are $F_1$ and $F_2$, then

$$P{F}_1+P{F}_2=\text{constant}.$$

This definition explains the geometric origin of the ellipse. However, for most H2 sketching questions, the more useful form is the standard coordinate form:

$$\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1.$$

This form should be read directly as geometry.

The centre is

$$(h,k).$$

The horizontal semi-axis length is

$$a.$$

The vertical semi-axis length is

$$b.$$

Here, $a$ means the horizontal distance from the centre to the ellipse, and $b$ means the vertical distance from the centre to the ellipse. It is not necessary that $a>b$.

From the centre $(h,k)$, the ellipse reaches the following four extreme points:

$$(h+a,k),(h-a,k),(h,k+b),(h,k-b).$$

So the full horizontal width is

$$2a,$$

and the full vertical height is

$$2b.$$

This is the key difference between a circle and an ellipse. For a circle, the distance from the centre to the curve is the same in every direction. For an ellipse, the horizontal and vertical distances are controlled separately.

For example, consider

$$\frac{(x-1{)}^2}{9}+\frac{(y+2{)}^2}{4}=1.$$

Since

$$9=3^2$$

and

$$4=2^2,$$

we read:

• centre: $(1,-2)$
• horizontal semi-axis length: $3$
• vertical semi-axis length: $2$

The four extreme points are

$$(4,-2),(-2,-2),(1,0),(1,-4).$$

These points are usually enough to guide a good sketch.

An ellipse has central symmetry about its centre. It is also symmetric about the horizontal and vertical lines through the centre:

$$y=k$$

and

$$x=h.$$

So the centre is the balancing point of the whole curve.

Caption: An ellipse is centred at $(h,k)$ and is determined by its horizontal and vertical semi-axis lengths.

What the algebra tells you

For a general quadratic equation without an $xy$ term,

$$A{x}^2+B{y}^2+Cx+Dy+E=0,$$

an ellipse is suggested when the coefficients of $x^2$ and $y^2$ have the same sign.

If those coefficients are equal, the curve may be a circle. If they have the same sign but are unequal, the curve may be a non-circular ellipse.

So the rough distinction is:

• equal same-sign coefficients suggest a circle
• unequal same-sign coefficients suggest an ellipse
• opposite-sign coefficients suggest a hyperbola

But this coefficient pattern is only a first clue. To confirm that the equation really represents an ellipse, we still need to complete the square and rewrite it into standard form.

For example, after completing the square, an ellipse should have the form

$$\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1,$$

where

$$a^2>0,b^2>0.$$

If the rearranged equation gives an impossible condition, then there is no real ellipse.

For instance,

$$\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=-1$$

cannot represent a real ellipse, because the left-hand side is always non-negative.

A circle can also be viewed as a special ellipse where the horizontal and vertical semi-axis lengths are equal.

What to identify

When sketching or interpreting an ellipse, identify:

• the centre
• the horizontal semi-axis length
• the vertical semi-axis length
• the four extreme points
• the axial intercepts, if they exist
• whether the ellipse just touches an axis

The centre gives the location of the ellipse.

The semi-axis lengths tell us how far the ellipse extends horizontally and vertically from the centre.

The four extreme points give a simple framework for drawing the curve.

Axial intercepts should be checked from the equation, not assumed from the sketch. A shifted ellipse may have no $x$-intercepts, no $y$-intercepts, or may just touch an axis.

For

$$\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1,$$

the ellipse just touches the $y$-axis when the horizontal distance from the centre to the $y$-axis equals the horizontal semi-axis length:

$$|h|=a.$$

It just touches the $x$-axis when the vertical distance from the centre to the $x$-axis equals the vertical semi-axis length:

$$|k|=b.$$

The useful habit is:

Do not only recognize the equation as an ellipse.
Read from the equation where the ellipse is centred, how far it extends, and how it sits relative to the axes.

Hyperbola

An ellipse is defined by a constant sum of distances from two fixed points. A hyperbola is defined by a constant difference of distances from two fixed points.

A hyperbola is the set of all points for which the absolute difference of distances from two fixed points is constant. These two fixed points are called the foci.

If $P(x,y)$ is a point on the hyperbola and the foci are $F_1$ and $F_2$, then

$$|P{F}_1-P{F}_2|=\text{constant}.$$

The absolute value is important because there are two branches. On one branch, points are closer to $F_1$ than to $F_2$. On the other branch, points are closer to $F_2$ than to $F_1$. This is one reason a hyperbola appears as two disconnected open curves rather than one closed loop.

For H2 sketching, the most useful forms are the two standard coordinate forms.

A hyperbola opening left and right has the form

$$\frac{(x-h{)}^2}{a^2}-\frac{(y-k{)}^2}{b^2}=1.$$

Its centre is

$$(h,k),$$

and its vertices are

$$(h-a,k),(h+a,k).$$

A hyperbola opening up and down has the form

$$\frac{(y-k{)}^2}{b^2}-\frac{(x-h{)}^2}{a^2}=1.$$

Its centre is again

$$(h,k),$$

and its vertices are

$$(h,k-b),(h,k+b).$$

The key reading rule is:

The positive squared term tells the direction in which the hyperbola opens.

So if the positive term contains $(x-h{)}^2$, the branches open left and right. If the positive term contains $(y-k{)}^2$, the branches open up and down.

This is the main structural difference from an ellipse. In an ellipse, the squared terms are added:

$$\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1.$$

In a hyperbola, one squared term is subtracted. That subtraction changes the curve from a closed loop into two open branches.

Asymptotes are part of the object

For a hyperbola, the asymptotes are not optional decoration. They are the guiding lines that control the long-range direction of the branches.

For both standard forms above, the asymptotes pass through the centre $(h,k)$ and have equations

$$y-k=\pm \frac{b}{a}(x-h).$$

The branches of the hyperbola approach these lines as they move farther away from the centre.

The constants $a$ and $b$ should therefore be read carefully:

• $a$ locates the vertices from the centre in the horizontal direction
• $b$ helps determine the slope of the asymptotes
• in the vertical-opening form, $b$ also locates the vertices vertically from the centre

This is slightly different from an ellipse. For an ellipse, $a$ and $b$ are semi-axis lengths. For a hyperbola, they help build the vertex positions and asymptote directions.

For example, consider

$$\frac{(x-1{)}^2}{9}-\frac{(y+2{)}^2}{4}=1.$$

Here,

$$h=1,k=-2,a=3,b=2.$$

Since the positive squared term is $(x-1{)}^2$, the hyperbola opens left and right.

The centre is

$$(1,-2).$$

The vertices are

$$(1-3,-2)=(-2,-2)$$

and

$$(1+3,-2)=(4,-2).$$

The asymptotes are

$$y+2=\pm \frac{2}{3}(x-1).$$

These features provide the framework for the sketch: mark the centre, mark the vertices, draw the dashed asymptotes, then draw two branches opening left and right while approaching the asymptotes.

Caption: A hyperbola is organised by its centre, vertices, and asymptotes, which determine the branch directions.

What the algebra tells you

For a general quadratic equation without an $xy$ term,

$$A{x}^2+B{y}^2+Cx+Dy+E=0,$$

a hyperbola is suggested when the coefficients of $x^2$ and $y^2$ have opposite signs.

For example,

$$4x^2-9y^2+8x+18y-41=0$$

has opposite signs on the squared terms, so it may represent a hyperbola.

But the coefficient pattern is only a first clue. Completing the square is still needed to reveal the centre, orientation, vertices, and asymptotes.

For instance, after completing the square, the equation should be rewritten into one of the standard forms:

$$\frac{(x-h{)}^2}{a^2}-\frac{(y-k{)}^2}{b^2}=1$$

or

$$\frac{(y-k{)}^2}{b^2}-\frac{(x-h{)}^2}{a^2}=1.$$

Only then can the sketch be drawn reliably.

What to identify

When sketching or interpreting a hyperbola, identify:

• the centre
• the direction of opening
• the vertices
• the asymptotes
• the axial intercepts, if they exist

The centre is where the asymptotes cross.

The direction of opening is determined by the positive squared term.

The vertices mark where the branches are closest to the centre.

The asymptotes give the long-range direction of the branches.

Axial intercepts should be checked from the equation, not inferred from the rough shape. A shifted hyperbola may have two intercepts, one intercept, or no intercepts on a given axis.

A reliable sketching workflow is:

1. Rewrite the equation into standard form if necessary.
2. Read the centre.
3. Decide whether the hyperbola opens horizontally or vertically.
4. Mark the vertices.
5. Draw the asymptotes as dashed lines.
6. Sketch the branches approaching the asymptotes.
7. Check axial intercepts if they are required.

The useful habit is:

For a hyperbola, do not draw the branches first.
Draw the asymptote framework first, then let the branches follow it.

Parabola

A parabola is different from the other conics in an important way.

A circle and an ellipse are closed curves. A hyperbola has two open branches. A parabola has one open branch with a single direction of opening.

For graph sketching, the essential facts are the vertex, axis of symmetry, and direction of opening.

As enrichment geometry, a parabola can also be described as the set of all points that are equally far from a fixed point and a fixed line.

The fixed point is called the focus.

The fixed line is called the directrix.

So a point $P(x,y)$ lies on the parabola when

$$\text{distance from }P\text{ to the focus}=\text{distance from }P\text{ to the directrix}.$$

The parabola sits halfway between the focus and the directrix. The point where the curve turns is called the vertex.

For the core graphing path, the vertex is the main organising feature of a parabola. Unlike a circle or an ellipse, a parabola has no centre. Unlike a hyperbola, it has no asymptotes. Its position and direction are controlled by its vertex and its axis of symmetry.

A parabola is therefore a vertex-driven conic.

There are two standard orientations.

A horizontally opening parabola has the form

$$(y-k{)}^2=4a(x-h).$$

Its vertex is

$$(h,k),$$

and its axis of symmetry is

$$y=k.$$

It opens to the right if

$$a>0,$$

and opens to the left if

$$a<0.$$

For enrichment, the focus is

$$(h+a,k),$$

and the directrix is

$$x=h-a.$$

A vertically opening parabola has the form

$$(x-h{)}^2=4a(y-k).$$

Its vertex is

$$(h,k),$$

and its axis of symmetry is

$$x=h.$$

It opens upwards if

$$a>0,$$

and opens downwards if

$$a<0.$$

For enrichment, the focus is

$$(h,k+a),$$

and the directrix is

$$y=k-a.$$

The sign of $a$ determines the direction of opening. In the focus-directrix interpretation, $|a|$ is the distance from the vertex to the focus and also from the vertex to the directrix.

For example, consider

$$(y+1{)}^2=8(x-2).$$

This matches the form

$$(y-k{)}^2=4a(x-h).$$

So

$$h=2,k=-1.$$

The vertex is therefore

$$(2,-1).$$

Also,

$$4a=8,$$

so

$$a=2.$$

Since $a>0$, the parabola opens to the right.

Its axis of symmetry is

$$y=-1.$$

For enrichment, its focus is

$$(h+a,k)=(4,-1),$$

and its directrix is

$$x=h-a=0.$$

This example shows how the equation gives the sketching information directly.

Caption: A parabola is organised for sketching by its vertex, axis, and direction of opening; the focus and directrix give the enrichment geometry.

What the algebra tells you

For a general quadratic equation without an $xy$ term,

$$A{x}^2+B{y}^2+Cx+Dy+E=0,$$

a parabola is suggested when exactly one of $A$ and $B$ is zero.

That means exactly one of $x^2$ and $y^2$ appears.

For example,

$$y^2-4y-8x+20=0$$

has a $y^2$ term but no $x^2$ term, so it may represent a horizontally opening parabola.

However, this coefficient pattern is only a first clue. Completing the square is still needed to reveal the vertex, axis of symmetry, and direction of opening.

For a genuine parabola in these standard forms, the other variable must appear linearly after rearrangement. The completed-square form is what confirms that the equation really behaves as a parabola rather than as a degenerate or impossible equation.

For instance,

$$y^2-4y-8x+20=0$$

can be rewritten as

$$(y-2{)}^2=8(x-2).$$

Now the structure is visible:

• vertex: $(2,2)$
• axis of symmetry: $y=2$
• opens to the right because $4a=8$, so $a=2>0$

A graph such as

$$y=x^2$$

is already familiar as a quadratic function. In this topic, however, we read it as a conic graph: through its vertex, symmetry, direction of opening, and standard form.

What to identify

When sketching or interpreting a parabola, identify:

• the vertex
• the axis of symmetry
• the direction of opening
• the focus and directrix, if required
• the axial intercepts, if they exist

The vertex is where the curve turns.

The axis of symmetry passes through the vertex. In the focus-directrix view, it also passes through the focus.

The direction of opening is determined by the standard form and the sign of $a$.

The intercepts should be solved from the equation, not guessed from a rough sketch.

A reliable sketching workflow is:

1. Rewrite the equation into standard form if necessary.
2. Identify the vertex.
3. Identify the axis of symmetry.
4. Use the sign of $a$ to decide the direction of opening.
5. Mark the focus and directrix if required.
6. Check axial intercepts if they are needed.
7. Sketch the single branch opening from the vertex.

The useful habit is:

For a parabola, do not look for a centre or asymptotes.
Find the vertex, then read the direction of opening.

Related

• Conic Sections
• Conic Classification and Sketching Workflow
• Graphing Techniques
• Content Map