Class 12 Mathematics

Conics

From definitions to tilted axes

The Core Idea

Every non-standard conic equation is a standard one with some transformations applied to the plane. There are only two operations we ever use:

Operation	What it does	How to apply it
Translation by $(h,k)$	Moves the origin to $(h,k)$	Replace $x \to x-h$, $y \to y-k$
Rotation by $\theta$	Tilts both axes by angle $\theta$	Substitute $x$ and $y$ using rotation formulas

That is the entire toolkit. Everything in this guide is just applying these two operations, one at a time, to a starting equation.

Why does rotation by 45° give $x = \dfrac{X-Y}{\sqrt{2}}$?

A point $P$ is fixed in the plane. We have two coordinate systems - the old axes $(x, y)$ and the new axes $(X, Y)$ rotated 45° counterclockwise. The point does not move; only the measuring frame changes.

Figure 1 - Same point P in old and new coordinate systems

The orange line is the new X-direction (along $y=x$); the blue line is the new Y-direction (along $y=-x$). The dashed segments are projections of $P$ onto each set of axes.

The unit vector along the new X-axis (which points along $y = x$) is:

$$\hat{X} = (\cos 45°,\ \sin 45°) = \left(\frac{1}{\sqrt{2}},\ \frac{1}{\sqrt{2}}\right)$$

The unit vector along the new Y-axis (perpendicular, 90° further) is:

$$\hat{Y} = (\cos 135°,\ \sin 135°) = \left(\frac{-1}{\sqrt{2}},\ \frac{1}{\sqrt{2}}\right)$$

The coordinate $X$ is the projection of $\overrightarrow{OP}$ onto $\hat{X}$ - that is, the dot product:

Derivation - projection onto rotated axes

$$X = \overrightarrow{OP} \cdot \hat{X} = x \cdot \frac{1}{\sqrt{2}} + y \cdot \frac{1}{\sqrt{2}}$$

dot product with $\hat{X}$

$$\boxed{X = \frac{x+y}{\sqrt{2}}}$$

simplified

$$Y = \overrightarrow{OP} \cdot \hat{Y} = x \cdot \frac{-1}{\sqrt{2}} + y \cdot \frac{1}{\sqrt{2}}$$

dot product with $\hat{Y}$

$$\boxed{Y = \frac{y-x}{\sqrt{2}}}$$

simplified

Now invert. We have two equations:

$$x + y = X\sqrt{2} \tag{1}$$ $$y - x = Y\sqrt{2} \tag{2}$$

Add $(1) + (2)$: $\;2y = (X+Y)\sqrt{2}$, so $y = \dfrac{X+Y}{\sqrt{2}}$.

Subtract $(2)$ from $(1)$: $\;2x = (X-Y)\sqrt{2}$, so:

Result - inverse rotation at 45° $$x = \frac{X-Y}{\sqrt{2}}, \qquad y = \frac{X+Y}{\sqrt{2}}$$

These come from solving two simultaneous linear equations - nothing more mysterious than that.

General angle θ

The same argument works for any angle. The new axes have unit vectors $(\cos\theta, \sin\theta)$ and $(-\sin\theta, \cos\theta)$, giving:

$$X = x\cos\theta + y\sin\theta, \qquad Y = -x\sin\theta + y\cos\theta$$

Inverting (solving simultaneously):

$$x = X\cos\theta - Y\sin\theta, \qquad y = X\sin\theta + Y\cos\theta$$

For $\theta = 45°$, $\cos\theta = \sin\theta = \frac{1}{\sqrt{2}}$, which recovers the 45° formulas exactly.

Figure 2 - Rotation by a general angle θ

The same projection argument works for any θ, not just 45°. Project $P$ onto the rotated X-axis (orange) and Y-axis (blue) to read off $X$ and $Y$ in terms of $x$, $y$, $\cos\theta$, $\sin\theta$.

Part I · The Parabola

The Parabola

Definition

A parabola is the set of all points $P$ in the plane that are equidistant from a fixed point $F$ (the focus) and a fixed line $\ell$ (the directrix).

$$PF = \text{distance from } P \text{ to } \ell$$

The axis is the line through $F$ perpendicular to $\ell$. The vertex is the midpoint between $F$ and $\ell$ - it lies on the axis.

Figure 3 - Parabola from its definition

Every point on the curve is exactly the same distance from the focus $F$ (red) as from the directrix line (blue). The vertex is halfway between them.

1.1 Standard equation - axis along $x$, vertex at origin

Place the vertex at the origin. The focus is at $F = (p, 0)$ and the directrix is $x = -p$. For any point $P = (x, y)$:

Full derivation from definition

$$PF = \text{dist to directrix}$$

definition

$$\sqrt{(x-p)^2 + y^2} = |x+p|$$

expanding both sides

$$(x-p)^2 + y^2 = (x+p)^2$$

square both sides

$$x^2 - 2px + p^2 + y^2 = x^2 + 2px + p^2$$

expand

$$y^2 = 4px$$

cancel $x^2, p^2$; collect

Standard equation - opens right $$y^2 = 4px$$

Vertex $(0,0)$. Focus $(p,0)$. Directrix $x = -p$. Axis: the $x$-axis.

Figure 4 - Standard parabola $y^2 = 4px$

Three curves show how the parabola widens as the focal parameter $p$ increases - the latus rectum (the focal chord) has length $4p$.

1.2 Translation - vertex at $(h, k)$

Shift the origin to $(h, k)$: replace $x \to x - h$ and $y \to y - k$.

Translated parabola - opens right $$(y-k)^2 = 4p(x-h)$$

Vertex $(h,k)$. Axis is the horizontal line $y = k$.

Figure 5 - Translated parabola $(y-k)^2 = 4p(x-h)$

The gray curve marks the original $y^2=4px$ at the origin; the orange curve is the same shape moved so its vertex sits at $(h,k)$. The axis is now the line $y=k$.

1.3 Rotation by 90° - axis along $y$, opens upward

Rotate the plane 90° counterclockwise. The 90° rotation substitution is $x \to y,\ y \to -x$. Apply to $y^2 = 4px$:

Derivation

Replace $y \to -x$, $x \to y$ in $y^2 = 4px$:

$$(-x)^2 = 4p(y)$$

substitution

$$x^2 = 4py$$

simplify

Translate vertex to $(h, k)$:

Parabola - opens upward $$(x-h)^2 = 4p(y-k)$$

Figure 6 - Parabola opening upward, axis along $y$

The gray curve is $x^2=4py$ at the origin; rotating 90° and translating gives the orange curve $(x-h)^2=4p(y-k)$, opening upward with vertex $(h,k)$ and focus $F$ directly above it.

1.4 Rotation by 180° - opens left and downward

Rotation by 180°: $x \to -x,\ y \to -y$. Apply to $y^2 = 4px$:

$$(-y)^2 = 4p(-x) \implies y^2 = -4px$$

Opens to the left. Translate to $(h,k)$:

Parabola - opens left $$(y-k)^2 = -4p(x-h)$$

Apply 180° to $x^2 = 4py$ instead: $x^2 = -4py$. Opens downward. Translate:

Parabola - opens downward $$(x-h)^2 = -4p(y-k)$$

Figure 7 - All four standard parabolas (your Class 12 syllabus)

Each is a rotation + translation of $y^2 = 4px$. The sign of $4p$ and whether $x$ or $y$ is squared tells you the direction.

These four forms - $(y-k)^2 = \pm4p(x-h)$ and $(x-h)^2 = \pm4p(y-k)$ - cover everything in the standard Class 12 syllabus. The next two sections go further.

1.5 Rotation by 45° - axis along $y = x$

Now rotate the plane by 45°. We established that the substitution to undo 45° rotation is $x = \dfrac{X-Y}{\sqrt{2}},\; y = \dfrac{X+Y}{\sqrt{2}}$, where $X, Y$ are the standard coordinates. So to rewrite the standard parabola $Y^2 = 4pX$ in old $(x,y)$ coordinates, substitute:

Derivation - 45° rotation of $Y^2 = 4pX$

$$\left(\frac{y-x}{\sqrt{2}}\right)^2 = 4p \cdot \frac{x+y}{\sqrt{2}}$$

substitute $X, Y$

$$\frac{(y-x)^2}{2} = \frac{4p(x+y)}{\sqrt{2}}$$

simplify left side

$$(y-x)^2 = \frac{2 \cdot 4p(x+y)}{\sqrt{2}} = 4p\sqrt{2}\,(x+y)$$

multiply both sides by 2

Parabola - axis along $y = x$, vertex at origin $$(y-x)^2 = 4p\sqrt{2}\,(x+y)$$

Translate vertex to $(h, k)$ - replace $x \to x-h$, $y \to y-k$:

Parabola - axis along $y = x$, vertex at $(h,k)$ $$\bigl(y - x - (k-h)\bigr)^2 = 4p\sqrt{2}\,\bigl(x + y - h - k\bigr)$$

Figure 8 - Parabola with axis along $y = x$

The axis of symmetry is the line $y = x$ (blue). The vertex sits at the origin. The $\sqrt{2}$ factor appears because the rotation stretches the coordinate in that direction.

1.6 General axis - $ax + by = c$

The line $ax + by = c$ has:

Normal direction $(a, b)$ - perpendicular to the line
Direction along the line: $(-b, a)$

So the two natural coordinate combinations are:

$$P = \frac{-bx + ay}{\sqrt{a^2+b^2}} \quad \text{(distance along the axis)}$$ $$Q = \frac{ax + by - c}{\sqrt{a^2+b^2}} \quad \text{(signed distance from the axis line)}$$

Notice that $Q$ is precisely the signed distance from the point $(x,y)$ to the line $ax+by=c$ - a formula you already know from coordinate geometry.

Parabola - general axis $ax+by=c$, vertex on the line $$Q^2 = 4p \cdot P$$ $$\left(\frac{ax+by-c}{\sqrt{a^2+b^2}}\right)^2 = 4p\left(\frac{-bx+ay}{\sqrt{a^2+b^2}}\right)$$

Check - the 45° case: with $a=1, b=-1, c=0$ (axis line $x-y=0$, i.e. $y=x$), $\sqrt{a^2+b^2}=\sqrt{2}$, so $P=\frac{x+y}{\sqrt2}$ and $Q=\frac{x-y}{\sqrt2}$. Substituting into $Q^2=4pP$ recovers $(y-x)^2 = 4p\sqrt{2}(x+y)$ exactly - matching the result derived above. ✓

Figure 9 - Parabola with a general axis $ax+by=c$

Here the axis is the line $x+2y=2$ (blue, dashed). $P$ measures position along the axis; $Q$ measures signed distance from it. Stand on the tilted axis and the curve $Q^2=4pP$ looks exactly like the standard parabola.

Part II · Ellipse & Circle

The Ellipse (and Circle)

Definition

An ellipse is the set of all points $P$ such that the sum of distances from $P$ to two fixed points $F_1$ and $F_2$ (the foci) is constant:

$$PF_1 + PF_2 = 2a_e \qquad (2a_e > F_1F_2)$$

Figure 10 - Ellipse from its definition

The sum $PF_1 + PF_2$ is constant for every point on the ellipse. Move $P$ mentally around the curve - the two distances change, but their sum stays at $2a_e$.

2.1 Standard equation - foci on $x$-axis, center at origin

Place foci at $(\pm c, 0)$, constant sum $= 2a_e$, with $a_e > c$. Define $b_e^2 = a_e^2 - c^2$.

Derivation

$$\sqrt{(x+c)^2+y^2} + \sqrt{(x-c)^2+y^2} = 2a_e$$

definition

$$\sqrt{(x+c)^2+y^2} = 2a_e - \sqrt{(x-c)^2+y^2}$$

isolate one radical

$$(x+c)^2+y^2 = 4a_e^2 - 4a_e\sqrt{(x-c)^2+y^2} + (x-c)^2+y^2$$

square both sides

$$4cx - 4a_e^2 = -4a_e\sqrt{(x-c)^2+y^2}$$

expand and simplify

$$a_e^2(x-c)^2 + a_e^2 y^2 = (a_e^2 - cx)^2$$

divide by $-4$, square again

$$(a_e^2 - c^2)x^2 + a_e^2 y^2 = a_e^2(a_e^2 - c^2)$$

expand and collect

$$\frac{x^2}{a_e^2} + \frac{y^2}{b_e^2} = 1$$

divide by $a_e^2 b_e^2$, use $b_e^2 = a_e^2 - c^2$

Standard ellipse - major axis along $x$ $$\frac{x^2}{a_e^2} + \frac{y^2}{b_e^2} = 1, \qquad a_e > b_e > 0$$

Center $(0,0)$. Foci $(\pm c, 0)$ where $c^2 = a_e^2 - b_e^2$. Semi-major axis $a_e$, semi-minor axis $b_e$.

Figure 11 - Standard ellipse, labelled $a$, $b$, $c$

Semi-major axis $a$ and semi-minor axis $b$ are measured from the center; the foci (red) sit at $(\pm c, 0)$ with $c^2 = a^2 - b^2$.

2.2 The circle emerges naturally

What happens if both foci coincide - i.e., $c = 0$? Then $b_e^2 = a_e^2 - 0 = a_e^2$, so $a_e = b_e = r$. The ellipse equation becomes:

$$\frac{x^2}{r^2} + \frac{y^2}{r^2} = 1 \implies x^2 + y^2 = r^2$$

Circle - special case of ellipse with $a_e = b_e$ $$x^2 + y^2 = r^2$$

The circle needs no separate definition - it is the ellipse whose two foci have merged into a single center. Every direction is a "major axis."

Figure 12 - Circle as limiting case of ellipse ($c \to 0$)

As $c$ shrinks toward $0$ (gray ellipses), the foci slide toward the center. At $c=0$ the curve becomes a perfect circle (orange) with its foci merged at one point.

2.3 Translation and 90° rotation

Translate center to $(h,k)$ - replace $x \to x-h$, $y \to y-k$:

Ellipse - center at $(h,k)$, major axis horizontal $$\frac{(x-h)^2}{a_e^2} + \frac{(y-k)^2}{b_e^2} = 1$$

For the circle: $(x-h)^2 + (y-k)^2 = r^2$.

Rotate 90° ($x \to y, y \to -x$) to get the major axis along $y$, then translate:

Ellipse - center at $(h,k)$, major axis vertical $$\frac{(x-h)^2}{b_e^2} + \frac{(y-k)^2}{a_e^2} = 1, \qquad a_e > b_e$$

The circle is unchanged by any rotation: $(x-h)^2 + (y-k)^2 = r^2$ looks identical after any rotation because both axes are equal - there is no preferred direction. This is the geometric reason circles have no axis.

Figure 13 - Ellipse after a 90° rotation and translation

The pale curve is the original ellipse at the origin; rotating 90° and translating gives the orange ellipse with a vertical major axis, centered at $(h,k)$.

2.4 Rotation by 45° - major axis along $y = x$

In the rotated frame the standard form is $\dfrac{X^2}{a_e^2} + \dfrac{Y^2}{b_e^2} = 1$. Substitute $X = \dfrac{x+y}{\sqrt{2}}$, $Y = \dfrac{y-x}{\sqrt{2}}$:

Derivation

$$\frac{1}{a_e^2}\left(\frac{x+y}{\sqrt{2}}\right)^2 + \frac{1}{b_e^2}\left(\frac{y-x}{\sqrt{2}}\right)^2 = 1$$

substitute $X, Y$

$$\frac{(x+y)^2}{2a_e^2} + \frac{(y-x)^2}{2b_e^2} = 1$$

simplify

Ellipse - major axis along $y = x$ $$\frac{(x+y)^2}{2a_e^2} + \frac{(y-x)^2}{2b_e^2} = 1$$

Check - circle is unchanged: Set $a_e = b_e = r$:

$$\frac{(x+y)^2 + (y-x)^2}{2r^2} = 1 \implies \frac{2x^2 + 2y^2}{2r^2} = 1 \implies x^2 + y^2 = r^2 \checkmark$$

Figure 14 - Ellipse with major axis along $y = x$

The ellipse is tilted 45°. Its major axis lies along the blue line $y=x$.

2.5 General axis - $ax + by = c$

Using the same $P$ (along axis) and $Q$ (perpendicular to axis) as for the parabola:

Ellipse - general axis $ax+by=c$ $$\frac{P^2}{a_e^2} + \frac{Q^2}{b_e^2} = 1$$

where $P = \dfrac{-bx+ay}{\sqrt{a^2+b^2}}$ and $Q = \dfrac{ax+by-c}{\sqrt{a^2+b^2}}$.

For the circle ($a_e = b_e = r$): $P^2 + Q^2 = r^2$, which expands to $(x-h)^2 + (y-k)^2 = r^2$ - the axis direction vanishes entirely.

Figure 15 - Ellipse with a general axis $ax+by=c$

Same idea as the parabola: $P$ runs along the tilted axis (blue, dashed), $Q$ measures perpendicular distance from it, and the foci (red) sit on the axis itself.

Part III · The Hyperbola

The Hyperbola

Definition

A hyperbola is the set of all points $P$ such that the difference of distances from $P$ to two fixed points $F_1$ and $F_2$ (the foci) is constant:

$$|PF_1 - PF_2| = 2a_h$$

Figure 16 - Hyperbola from its definition

The difference (not sum) of distances to the two foci is constant. This gives two separate branches - one where $PF_1 - PF_2 = 2a_h$ and one where $PF_2 - PF_1 = 2a_h$.

3.1 Standard equation - foci on $x$-axis, center at origin

Place foci at $(\pm c, 0)$, constant difference $2a_h$, with $c > a_h$. Define $b_h^2 = c^2 - a_h^2$ (note: reversed from the ellipse).

Derivation (parallel to ellipse, but difference)

$$\left|\sqrt{(x+c)^2+y^2} - \sqrt{(x-c)^2+y^2}\right| = 2a_h$$

definition

Isolate one radical, square, isolate again, square:

$$(c^2 - a_h^2)x^2 - a_h^2 y^2 = a_h^2(c^2 - a_h^2)$$

after double squaring (same algebra as ellipse, sign flips)

$$\frac{x^2}{a_h^2} - \frac{y^2}{b_h^2} = 1$$

divide by $a_h^2 b_h^2$, use $b_h^2 = c^2 - a_h^2$

Standard hyperbola - transverse axis along $x$ $$\frac{x^2}{a_h^2} - \frac{y^2}{b_h^2} = 1$$

Center $(0,0)$. Foci $(\pm c, 0)$ where $c^2 = a_h^2 + b_h^2$. Asymptotes $y = \pm\frac{b_h}{a_h}x$. Vertices $(\pm a_h, 0)$.

Translate center to $(h, k)$:

Hyperbola - transverse axis horizontal, center $(h,k)$ $$\frac{(x-h)^2}{a_h^2} - \frac{(y-k)^2}{b_h^2} = 1$$

Figure 17 - Standard hyperbola, labelled $a$, $b$, $c$

Vertices sit at $(\pm a, 0)$; the asymptotes $y=\pm(b/a)x$ (blue, dashed) bound the branches; the foci (red) sit further out at $(\pm c, 0)$ with $c^2=a^2+b^2$.

3.2 Rotation by 90° - transverse axis along $y$

Rotate 90° ($x \to y, y \to -x$) and translate:

Hyperbola - transverse axis vertical $$\frac{(y-k)^2}{a_h^2} - \frac{(x-h)^2}{b_h^2} = 1$$

Note the sign: $y^2$ is now positive. The branches open up and down.

Figure 18 - Hyperbola in both orientations

Left (orange): transverse axis horizontal, branches open left-right. Right (blue): transverse axis vertical, branches open up-down.

3.3 Rotation by 45° - rectangular hyperbola $xy = k$

Rotate $\dfrac{X^2}{a_h^2} - \dfrac{Y^2}{b_h^2} = 1$ by 45°. Substitute $X = \dfrac{x+y}{\sqrt{2}}$, $Y = \dfrac{y-x}{\sqrt{2}}$:

Hyperbola - axis along $y = x$ $$\frac{(x+y)^2}{2a_h^2} - \frac{(y-x)^2}{2b_h^2} = 1$$

Special case - rectangular hyperbola ($a_h = b_h$):

Derivation of $xy = k$

$$\frac{(x+y)^2 - (y-x)^2}{2a_h^2} = 1$$

set $a_h = b_h$

$$\frac{4xy}{2a_h^2} = 1$$

$(x+y)^2 - (y-x)^2 = 4xy$

$$xy = \frac{a_h^2}{2}$$

i.e. $xy = k$

Rectangular hyperbola - asymptotes along $x$ and $y$ axes $$xy = k$$

This is simply $x^2 - y^2 = 2k$ rotated 45°. Its asymptotes are the coordinate axes. You see this equation in physics (Boyle's law $PV = \text{const}$) and economics (demand curves).

Figure 19 - Rectangular hyperbola $xy = k$ and its parent $x^2 - y^2 = \text{const}$

The rectangular hyperbola $xy = k$ (orange) is exactly $x^2 - y^2 = \text{const}$ (blue) rotated by 45°. The coordinate axes are the asymptotes.

3.4 General axis - $ax + by = c$

Hyperbola - general axis $ax+by=c$ $$\frac{P^2}{a_h^2} - \frac{Q^2}{b_h^2} = 1$$

where $P = \dfrac{-bx+ay}{\sqrt{a^2+b^2}}$ and $Q = \dfrac{ax+by-c}{\sqrt{a^2+b^2}}$.

Figure 20 - Hyperbola with a general axis $ax+by=c$

The same $P, Q$ construction as before, now producing two branches that open along the tilted axis (blue, dashed) instead of the $x$-axis.

Summary

The Grand Unified Picture

Every conic with axis along $ax + by = c$ is described by just two quantities:

$$P = \frac{-bx+ay}{\sqrt{a^2+b^2}} \quad \text{(position along the axis)}$$ $$Q = \frac{ax+by-c}{\sqrt{a^2+b^2}} \quad \text{(signed distance from the axis line)}$$

Conic	Equation in $P, Q$	What it says geometrically
Parabola	$Q^2 = 4pP$	Distance-to-axis² grows linearly along the axis
Ellipse	$\dfrac{P^2}{\alpha^2} + \dfrac{Q^2}{\beta^2} = 1$	Bounded in both directions
Circle	$P^2 + Q^2 = r^2$	Equal bound in all directions - axis vanishes
Hyperbola	$\dfrac{P^2}{\alpha^2} - \dfrac{Q^2}{\beta^2} = 1$	Unbounded, opens along the axis - two branches

Figure 21 - All four conics on one set of axes

Parabola (orange), ellipse (blue), circle (green), hyperbola (red) - four equations, one substitution pattern.

And the journey to get here was just a single pipeline applied repeatedly:

Definition $\xrightarrow{\text{algebra}}$ Standard form $\xrightarrow{+\,(h,k)}$ Translated $\xrightarrow{+\,\theta}$ Rotated $\xrightarrow{\text{algebra}}$ General