Conic Sections

Slice a cone with a plane and, depending only on the angle of the cut, you get a circle, an ellipse, a parabola, or a hyperbola — four curves that look wildly different on paper but are secretly one family, born from the same solid. The ancient Greek geometer Apollonius named two of them, parabola and hyperbola, more than two thousand years ago, and the family has never lost its usefulness: planets trace ellipses around the Sun, thrown balls and reflector dishes trace parabolas, and antennas and flashlight mirrors are shaped from these same curves. This chapter builds each curve from a clean point-set definition rather than from cutting cones, then hunts for the simplest possible equation in Cartesian coordinates. The method is the same every time. Fix one or two special points, a focus or a pair of foci, and sometimes a line, a directrix. Describe the curve as the set of points obeying one geometric rule about distances to those points. Place the coordinate axes cleverly so the algebra collapses to something short, and derive the equation using nothing more than the distance formula and patient squaring. A circle is the set of points at a fixed distance from a centre. A parabola balances distance to a point against distance to a line. An ellipse fixes the sum of two distances; a hyperbola fixes their difference. Four definitions, four short derivations, four standard equations, and one number, the eccentricity, that secretly measures how un-circular each curve is. By the end you should be able to go in either direction: given a geometric description such as a focus, a vertex, a directrix, or an eccentricity, write down the equation, and given an equation, read off every geometric feature, the centre, foci, vertices, axes, eccentricity and latus rectum, at a glance. That two-way fluency is exactly what CBSE boards and JEE or NEET test, often within the same question.

Key takeaways

A conic section is the curve formed when a plane cuts a double napped right circular cone; the angle of the cut determines whether the section is a circle, ellipse, parabola or hyperbola.
A circle is the set of points at a fixed distance r, the radius, from a fixed centre (h, k); its equation is $(x-h)^2+(y-k)^2=r^2$ .
A parabola is the set of points equidistant from a fixed focus and a fixed directrix; the standard form $y^2=4ax$ has focus $(a,0)$ , directrix $x=-a$ , and latus rectum of length $4a$ .
An ellipse is the set of points whose distances from two foci add to a constant $2a$ ; the standard form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ (with $a\gt b$ ) satisfies $a^2=b^2+c^2$ , eccentricity $e=c/a\lt 1$ .
A hyperbola is the set of points whose distances from two foci differ by a constant $2a$ ; the standard form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ satisfies $b^2=c^2-a^2$ , eccentricity $e=c/a\gt 1$ .
Latus rectum length is $4a$ for a parabola but $2b^2/a$ for both the ellipse and the hyperbola, a formula pair students frequently mix up.
Eccentricity is the single number that classifies every conic: $e=0$ is a circle (the limiting ellipse), $0\lt e\lt 1$ an ellipse, $e=1$ a parabola, $e\gt 1$ a hyperbola.
When the intersecting plane passes through the vertex of the cone, the section degenerates to a point, a straight line, or a pair of intersecting lines.
Which axis of an ellipse is the major axis is decided by comparing the two denominators, never by which variable is squared first.
Key idea: every conic section is the locus of points satisfying one distance rule about a focus (or foci) and, for the parabola, a directrix, and placing the coordinate axes along the curve's own symmetry turns that geometric rule into one of four short standard equations.

The core insight

A conic section is the locus of points obeying a single distance rule involving a focus (or a pair of foci) and, for the parabola, a directrix; choosing axes along the curve's symmetry turns that rule into one of exactly four standard algebraic equations, distinguished by the eccentricity $e=0$ , $0\lt e\lt 1$ , $e=1$ and $e\gt 1$ for the circle, ellipse, parabola and hyperbola respectively.

What this chapter covers

This chapter opens with conic sections as slices of a cone: the vertex, axis and generators of a double napped cone, the angle conditions that produce a circle, ellipse, parabola or hyperbola, and the degenerate cases when the cutting plane passes through the vertex. It then treats each curve in turn purely as a locus in the plane, with no cones required. The circle is defined by a fixed distance from a centre. The parabola is defined by equal distances to a focus and a directrix, giving four standard orientations and the latus rectum. The ellipse and hyperbola are defined by a fixed sum, respectively difference, of distances to two foci, each with its own standard equation, eccentricity, and latus rectum. Three full derivations rebuild the standard equations from their locus definitions, three worked examples show the geometry-to-equation and equation-to-geometry conversions in both directions, and the chapter closes with eccentricity as the single number that unifies all four curves.

The core ideas

Conic sections as slices of a cone

Take a fixed vertical line $l$ and a second line $m$ that meets it at a fixed point $V$ , inclined to it at a constant angle $\alpha$ . Rotating $m$ about $l$ while keeping $\alpha$ fixed sweeps out a double napped right circular cone: $V$ is the vertex, $l$ is the axis, $m$ in any of its positions is a generator, and the vertex separates the surface into two nappes extending indefinitely in both directions. A conic section is simply the curve obtained by intersecting this cone with a plane, and which curve you get depends only on the angle $\beta$ that the plane makes with the axis. When the plane cuts a nappe away from the vertex, four things can happen. If $\beta=90^\circ$ the plane is perpendicular to the axis and the section is a circle. If $\alpha\lt \beta\lt 90^\circ$ the section is an ellipse. If $\beta=\alpha$ , so the plane is parallel to a generator, the section is a parabola. If $0\le\beta\lt \alpha$ the plane is steep enough to cut through both nappes, and the section is a hyperbola, appearing as two separate branches. If instead the plane passes through the vertex itself, the same four angle ranges give degenerate conics: a single point (when $\alpha\lt \beta\le90^\circ$ ), a single straight line lying along a generator (when $\beta=\alpha$ , the degenerate parabola), or a pair of intersecting straight lines (when $0\le\beta\lt \alpha$ , the degenerate hyperbola).

Figure

Four schematic double napped cones each sliced by a plane at a different angle producing a circle an ellipse a parabola and a hyperbola with the vertex V and axis l marked in every panel

The circle

A circle is the set of all points in a plane equidistant from a fixed point. The fixed point is the centre and the constant distance is the radius $r$ . If the centre is $C(h,k)$ and $P(x,y)$ is any point on the circle, the defining condition $|CP|=r$ becomes, by the distance formula, the standard equation

(x-h)^2+(y-k)^2=r^2. \qquad (1)

where: $(h,k)$ is the centre and $r\gt 0$ is the radius; when the centre is the origin this collapses to the simplest form $x^2+y^2=r^2$ . Expanding equation (1) always produces a general equation of the form $x^2+y^2+2gx+2fy+c=0$ , and reversing that expansion by completing the square recovers centre $(-g,-f)$ and radius $r=\sqrt{g^2+f^2-c}$ , which is real only when $g^2+f^2-c\ge0$ . This general form is how a circle usually appears disguised in a problem, and completing the square is always the way back to equation (1).

The parabola

A parabola is the set of all points in a plane equidistant from a fixed line, the directrix, and a fixed point not on that line, the focus. The line through the focus perpendicular to the directrix is the axis of the parabola, and the point where the parabola meets its axis is the vertex. Placing the vertex at the origin with the focus at $(a,0)$ , $a\gt 0$ , and the directrix as the line $x=-a$ gives the simplest orientation, whose equation is derived in full below:

y^2=4ax. \qquad (2)

where: $a\gt 0$ is the distance from vertex to focus, equal to half the distance from focus to directrix. The three remaining orientations follow from reflecting this construction: $y^2=-4ax$ opens to the left with focus $(-a,0)$ and directrix $x=a$ ; $x^2=4ay$ opens upward with focus $(0,a)$ and directrix $y=-a$ ; $x^2=-4ay$ opens downward with focus $(0,-a)$ and directrix $y=a$ . In every case, whichever variable appears squared fixes the axis of symmetry (a $y^2$ term means the axis is the x-axis, an $x^2$ term means the axis is the y-axis), and the sign of the linear term fixes the direction the curve opens. The latus rectum is the chord through the focus perpendicular to the axis, with endpoints on the curve; substituting $x=a$ into equation (2) gives $y=\pm2a$ , so the length of the latus rectum is $4a$ .

The ellipse

An ellipse is the set of all points in a plane the sum of whose distances from two fixed points, the foci, is a constant. The midpoint of the segment joining the foci is the centre; the line through the foci is the major axis, and the perpendicular line through the centre is the minor axis. Writing the length of the major axis as $2a$ , the minor axis as $2b$ , and the distance between the foci as $2c$ , a point $P$ at the end of the major axis gives sum of distances $(c+a)+(a-c)=2a$ , while a point $Q$ at the end of the minor axis gives sum of distances $2\sqrt{b^2+c^2}$ ; equating these two expressions for the same constant, $2\sqrt{b^2+c^2}=2a$ , yields the fundamental relation of the ellipse:

a^2=b^2+c^2, \qquad\text{so}\qquad c=\sqrt{a^2-b^2}. \qquad (3)

where: $a$ is the semi-major axis, $b$ the semi-minor axis, and $c$ the centre-to-focus distance, with $a$ always the largest of the three. Placing the centre at the origin with foci on the x-axis at $(\pm c,0)$ gives the standard equation, derived in full below,

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \qquad a\gt b\gt 0. \qquad (4)

where: the major axis lies along whichever coordinate axis carries the larger denominator, regardless of whether that denominator sits under $x^2$ or $y^2$ ; swapping the roles of $a^2$ and $b^2$ gives the equation $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ for foci on the y-axis. The eccentricity is $e=c/a$ , always strictly between $0$ and $1$ because $a\gt c$ , and the latus rectum, the focal chord perpendicular to the major axis, has length $2b^2/a$ .

The hyperbola

A hyperbola is the set of all points in a plane the difference of whose distances from two fixed foci is a constant, where difference always means the farther distance minus the nearer one. The line through the foci is the transverse axis and the perpendicular line through the centre is the conjugate axis; the curve meets the transverse axis at two vertices, a distance $2a$ apart, while the foci are a distance $2c$ apart with $c\gt a$ . Unlike the ellipse, the defining quantity $b$ is introduced by the opposite subtraction,

b^2=c^2-a^2. \qquad (5)

where: $2b$ is the length of the conjugate axis, and because $c\gt a$ always, $b^2$ is automatically positive with no restriction relating $b$ to $a$ ; a hyperbola with $a=b$ is called equilateral, with eccentricity $e=\sqrt2$ . With foci on the x-axis at $(\pm c,0)$ , the standard equation, derived in full below, is

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1. \qquad (6)

where: the transverse axis lies along whichever coordinate axis carries the positive squared term; the mirror form $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ has its transverse axis on the y-axis. Because $x^2/a^2=1+y^2/b^2\ge1$ , every point satisfies $x\le-a$ or $x\ge a$ , so the curve never enters the strip between the lines $x=-a$ and $x=a$ . The eccentricity $e=c/a$ is always strictly greater than $1$ , and the latus rectum again has length $2b^2/a$ .

Eccentricity: one number that classifies every conic

Eccentricity is the ratio $e$ of the distance from the centre to a focus and the distance from the centre to a vertex, and for the parabola, ellipse and hyperbola it also equals the constant ratio of a point's distance from the focus to its distance from the corresponding directrix, $SP=e\cdot PM$ . As $e$ increases from $0$ , the family of conics sweeps through every shape: $e=0$ is the circle, the limiting case of an ellipse in which the two foci merge into a single centre and $a=b=r$ (its directrix having receded to infinity); $0\lt e\lt 1$ gives every ellipse in between; $e=1$ is exactly the parabola, the balance point where distance to focus equals distance to directrix outright; and $e\gt 1$ gives every hyperbola, growing ever flatter and more open as $e$ increases further. This one number is the fastest way to identify a conic without even solving for its other features.

Figure

An ellipse with semi major axis five semi minor axis three and foci at plus or minus four on the x axis showing the vertices the minor axis endpoints a focal chord latus rectum and a point P whose distances to the two foci sum to ten

Figure

Four standard parabolas in a two by two grid opening right left up and down each showing the vertex the focus the directrix and its equation

Derivation AStandard equation of a parabola

The goal here is to convert the pure geometric definition of a parabola, equal distance from a fixed focus and a fixed directrix, into a Cartesian equation, and to do it in the coordinate system that makes the algebra shortest: axes chosen along the parabola's own axis of symmetry rather than left in some arbitrary orientation. We derive the case with focus to the right of the origin and directrix to the left, then read off the other three orientations by symmetry without repeating the algebra.

Step 1. Set up coordinates. Let $F$ be the focus and $l$ the directrix, and let $M$ be the foot of the perpendicular from $F$ to $l$ . Let $O$ bisect $FM$ ; by the very definition of the parabola, $O$ is equidistant from $F$ and $l$ , so $O$ itself lies on the curve and is its vertex. Take $O$ as the origin, the ray $OF$ as the positive x-axis, and the perpendicular through $O$ as the y-axis. If the distance from directrix to focus is taken as $2a$ , the focus is $F(a,0)$ and the directrix is the line $x=-a$ .

Step 2. Translate the defining rule into distances. Let $P(x,y)$ be any point on the parabola and let $B$ be the foot of the perpendicular from $P$ to the directrix, so $B=(-a,y)$ . The defining property is $PF=PB$ . By the distance formula, $PF=\sqrt{(x-a)^2+y^2}$ and $PB=\sqrt{(x+a)^2}$ .

Step 3. Square and simplify. Setting $PF=PB$ and squaring both sides removes the radicals: $(x-a)^2+y^2=(x+a)^2$ . Expanding, $x^2-2ax+a^2+y^2=x^2+2ax+a^2$ , and the $x^2$ and $a^2$ terms cancel from both sides, leaving $y^2=4ax$ , with $a\gt 0$ . Every point on the parabola therefore satisfies this equation.

Step 4. Confirm the converse. For the equation to describe exactly the same set of points as the geometric definition, any point satisfying $y^2=4ax$ must also satisfy $PF=PB$ . Substituting $y^2=4ax$ into $PF$ gives $PF=\sqrt{(x-a)^2+4ax}=\sqrt{x^2+2ax+a^2}=\sqrt{(x+a)^2}=PB$ , since $x\ge0$ for points on the curve, making $x+a\gt 0$ . So the equation and the geometric definition describe exactly the same locus, and $y^2=4ax$ is the standard equation of the parabola with vertex at the origin, focus $(a,0)$ , and directrix $x=-a$ . Reflecting the construction across the axes or exchanging the roles of $x$ and $y$ gives, without repeating the algebra, the other three standard forms $y^2=-4ax$ , $x^2=4ay$ and $x^2=-4ay$ . Verified: substituting $x=y^2/4a$ back into $PF-PB$ gives zero identically, and setting $x=a$ in $y^2=4ax$ gives the latus-rectum endpoints $y=\pm2a$ , a chord of length $4a$ , matching the definition of the latus rectum as the focal chord perpendicular to the axis.

Go deeper · Derivation B · 2 minStandard equation of an ellipse

Derivation BStandard equation of an ellipse

The task is to convert the two-focus, sum-of-distances definition of an ellipse into a Cartesian equation, again by choosing axes along the curve's own symmetry so the two square roots that the distance formula produces can be cleared systematically. The derivation needs two rounds of squaring, because two separate radicals appear, and the whole trick is organising the algebra so each squaring removes exactly one radical rather than tangling both together.

Step 1. Set up coordinates and the defining sum. Let $F_1$ and $F_2$ be the foci and $O$ their midpoint. Take $O$ as the origin and the ray from $O$ through $F_2$ as the positive x-axis, so $F_1=(-c,0)$ and $F_2=(c,0)$ . Let $P(x,y)$ be a point on the ellipse with $PF_1+PF_2=2a$ , a fixed constant greater than $2c$ .

Step 2. Write the distances and isolate one radical. By the distance formula, $\sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}=2a$ . Moving the second radical to the right gives $\sqrt{(x+c)^2+y^2}=2a-\sqrt{(x-c)^2+y^2}$ , ready for the first squaring.

Step 3. Square once. Squaring both sides gives $(x+c)^2+y^2=4a^2-4a\sqrt{(x-c)^2+y^2}+(x-c)^2+y^2$ . The $y^2$ terms cancel, and since $(x+c)^2-(x-c)^2=4cx$ , collecting terms and dividing by $4a$ reduces this to a single remaining radical: $\sqrt{(x-c)^2+y^2}=a-\frac{c}{a}x$ .

Step 4. Square again and use $c^2=a^2-b^2$ . Squaring once more, $(x-c)^2+y^2=a^2-2cx+\frac{c^2}{a^2}x^2$ . Expanding the left side and cancelling the common $-2cx$ term from both sides, then multiplying through by $a^2$ and collecting the $x^2$ terms, leaves $x^2(a^2-c^2)+a^2y^2=a^2(a^2-c^2)$ . Writing $b^2=a^2-c^2$ , a positive quantity because $2a\gt 2c$ , and dividing throughout by $a^2b^2$ gives $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ .

Step 5. Confirm the converse. Substituting back, a point satisfying $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ gives $PF_1=a+\frac{c}{a}x$ and $PF_2=a-\frac{c}{a}x$ , each obtained by the same simplification using $b^2=a^2-c^2$ inside the square root. Adding these two expressions cancels the $\frac{c}{a}x$ terms exactly, returning $PF_1+PF_2=2a$ . The two-way equivalence therefore holds, and $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ , with $a\gt b\gt 0$ and $c^2=a^2-b^2$ , is the standard equation of the ellipse with foci $(\pm c,0)$ . The mirror construction with foci on the y-axis gives $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ by exactly the same steps with $x$ and $y$ exchanged. Verified: adding the back-substituted $PF_1$ and $PF_2$ gives exactly $2a$ with no leftover terms, and setting $x=c$ recovers the latus-rectum half-length $l=b^2/a$ , matching the length $2b^2/a$ derived independently in the core-ideas section.

Go deeper · Derivation C · 2 minStandard equation of a hyperbola

Derivation CStandard equation of a hyperbola

The goal is to run the same distance-formula method used for the ellipse, but now with a difference of distances fixed instead of a sum, and to show precisely where the algebra forks away from the ellipse case, producing the opposite sign in front of $y^2$ and the opposite relation between $a$ , $b$ and $c$ . As with the ellipse, two squarings are needed to clear two radicals in turn.

Step 1. Set up coordinates and the defining difference. With foci $F_1(-c,0)$ and $F_2(c,0)$ as before, let $P(x,y)$ be a point on the branch nearer $F_2$ , satisfying $PF_1-PF_2=2a$ (the farther distance minus the nearer one).

Step 2. Write the distances and isolate one radical. $\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}=2a$ , so $\sqrt{(x+c)^2+y^2}=2a+\sqrt{(x-c)^2+y^2}$ .

Step 3. Square once. $(x+c)^2+y^2=4a^2+4a\sqrt{(x-c)^2+y^2}+(x-c)^2+y^2$ . As before, the $y^2$ terms cancel and $(x+c)^2-(x-c)^2=4cx$ ; collecting terms and dividing by $4a$ gives $\sqrt{(x-c)^2+y^2}=\frac{c}{a}x-a$ , the same shape as the ellipse's Step 3 but with the sign of $a$ flipped.

Step 4. Square again and use $b^2=c^2-a^2$ . Squaring gives $(x-c)^2+y^2=\frac{c^2}{a^2}x^2-2cx+a^2$ . Expanding the left side and cancelling the common $-2cx$ term, then multiplying through by $a^2$ and collecting the $x^2$ terms, leaves $x^2(c^2-a^2)-a^2y^2=a^2(c^2-a^2)$ . Writing $b^2=c^2-a^2$ , positive since $c\gt a$ for a hyperbola, and dividing by $a^2b^2$ gives $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ .

Step 5. Confirm the converse and note the sign flip against the ellipse. Back-substitution gives $PF_1=a+\frac{c}{a}x$ and $PF_2=\frac{c}{a}x-a$ for points with $x\ge a$ , so $PF_1-PF_2=2a$ exactly; a symmetric argument on the branch with $x\le-a$ gives $PF_2-PF_1=2a$ , matching the farther-minus-nearer convention on that branch too. So $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ , with $b^2=c^2-a^2$ , is the standard hyperbola with foci $(\pm c,0)$ . Contrast this directly with the ellipse: there $b^2=a^2-c^2$ , subtracting the focal distance from the semi-major axis, while here $b^2=c^2-a^2$ , subtracting the semi-transverse axis from the focal distance. That reversal is exactly why the two curves look so different, and why the ellipse's $b$ is always smaller than $a$ while the hyperbola's $b$ carries no such restriction at all. Verified: on the branch $x\ge a$ , $PF_1-PF_2=\left(a+\frac{c}{a}x\right)-\left(\frac{c}{a}x-a\right)=2a$ identically, independent of $x$ , and setting $x=c$ recovers the latus-rectum half-length $b^2/a$ , matching $2b^2/a$ used in the core-ideas section.

Worked examples

Board level - centre and radius of a circle from the general equation

Find the centre and radius of the circle $x^2+y^2+8x+10y-8=0$ , and verify using the general-form shortcut.

Step 1. Group and complete the square. Rewrite as $(x^2+8x)+(y^2+10y)=8$ . Completing the square in each bracket, $(x^2+8x+16)+(y^2+10y+25)=8+16+25$ , which is $(x+4)^2+(y+5)^2=49$ .

Step 2. Read off the centre and radius. Comparing with $(x-h)^2+(y-k)^2=r^2$ , written as $\{x-(-4)\}^2+\{y-(-5)\}^2=7^2$ , the centre is $(-4,-5)$ and the radius is $7$ .

Step 3. Cross-check with the general-form shortcut. Matching $x^2+y^2+2gx+2fy+c=0$ to the given equation gives $2g=8$ so $g=4$ , $2f=10$ so $f=5$ , and $c=-8$ . The shortcut formulas centre $=(-g,-f)$ and radius $=\sqrt{g^2+f^2-c}$ give centre $(-4,-5)$ and radius $\sqrt{16+25+8}=\sqrt{49}=7$ , matching Step 2 exactly. Answer: centre $(-4,-5)$ , radius $7$ .

Board level - foci, vertices, axes, eccentricity and latus rectum of an ellipse

For the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ , find the coordinates of the foci and vertices, the lengths of the major and minor axes, the eccentricity, and the length of the latus rectum.

Step 1. Identify the major axis. The denominator under $x^2$ is $25$ and under $y^2$ is $9$ ; since $25\gt 9$ , the major axis lies along the x-axis, so the standard form is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a^2=25$ , $b^2=9$ , that is $a=5$ , $b=3$ .

Step 2. Find $c$ from $a^2=b^2+c^2$ . $c=\sqrt{a^2-b^2}=\sqrt{25-9}=\sqrt{16}=4$ . The foci are $(\pm4,0)$ and the vertices are $(\pm5,0)$ .

Step 3. Compute the remaining quantities. Length of major axis $=2a=10$ ; length of minor axis $=2b=6$ ; eccentricity $e=c/a=4/5=0.8$ ; latus rectum $=2b^2/a=2(9)/5=18/5=3.6$ . Answer: foci $(\pm4,0)$ , vertices $(\pm5,0)$ , major axis $10$ , minor axis $6$ , $e=4/5$ , latus rectum $=18/5$ .

Competitive level (JEE / NEET) - hyperbola from foci and latus rectum, with an extraneous root

A hyperbola has its foci on the y-axis at $(0,\pm12)$ , and the length of its latus rectum is $36$ . Find its equation.

Step 1. Translate the data. Foci on the y-axis at $(0,\pm c)$ means $c=12$ and the standard form is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ . The latus rectum formula $2b^2/a=36$ gives $b^2=18a$ .

Step 2. Substitute into $c^2=a^2+b^2$ . $144=a^2+18a$ , that is $a^2+18a-144=0$ , a quadratic in $a$ . Solving with the quadratic formula, $a=\frac{-18\pm\sqrt{324+576}}{2}=\frac{-18\pm\sqrt{900}}{2}=\frac{-18\pm30}{2}$ , giving $a=6$ or $a=-24$ .

Step 3. Reject the impossible root and finish. A length $a$ must be positive, so $a=-24$ is rejected outright; a common JEE trap is carrying both roots through to the final equation instead of discarding the negative one immediately. Taking $a=6$ gives $b^2=18(6)=108$ . Answer: the equation is $\frac{y^2}{36}-\frac{x^2}{108}=1$ , equivalently $3y^2-x^2=108$ . Timing note: this should take under two minutes once the latus rectum formula is remembered correctly as $2b^2/a$ , not $2a^2/b$ , because everything after that is one substitution and one quadratic. The real trap is arithmetic and sign discipline in the quadratic formula, plus the discipline to discard the negative root for $a$ on sight rather than plugging both roots back into the original relation to check.

Common traps

Deciding the major axis of an ellipse by which variable "comes first" instead of comparing denominators: whichever denominator is larger fixes the major axis, regardless of whether it sits under $x^2$ or $y^2$ .
Swapping the sign convention in $b^2$ between the ellipse and the hyperbola: for an ellipse $b^2=a^2-c^2$ , so $b\lt a$ always, but for a hyperbola $b^2=c^2-a^2$ , so $b$ can be anything positive; using the wrong subtraction order gives an impossible negative square.
Using the parabola's latus rectum formula $4a$ for an ellipse or hyperbola, or vice versa; the correct formula for the ellipse and hyperbola alike is $2b^2/a$ .
Forgetting the sign of the directrix relative to the focus: for $y^2=4ax$ opening right with focus $(a,0)$ , $a\gt 0$ , the directrix is the line $x=-a$ , not $x=a$ .
Assuming eccentricity can never exceed $1$ : that is true only for the ellipse ( $0\lt e\lt 1$ ); a hyperbola always has $e\gt 1$ , and treating a hyperbola's $e$ as though it must be a proper fraction is a common slip.
Getting the sign of the centre backward when converting a circle's general equation $x^2+y^2+2gx+2fy+c=0$ to centre-radius form: the centre is $(-g,-f)$ , not $(g,f)$ .
Treating the hyperbola's "difference of distances" as a signed subtraction rather than farther-minus-nearer, which can lead to a negative constant on one branch if the order is not fixed carefully.
Reporting $r^2$ , $a^2$ , $b^2$ or $c^2$ as though they were $r$ , $a$ , $b$ or $c$ themselves when substituting into eccentricity or latus rectum formulas, instead of taking the square root first.

Common trapThe ellipse relation a squared equals b squared plus c squared and the hyperbola relation b squared equals c squared minus a squared point in opposite directions, and mixing them up is the single most common error in this chapter

For an ellipse, the semi-major axis is the largest of the three lengths, so $a^2=b^2+c^2$ , hence $c^2=a^2-b^2$ : you subtract the smaller semi-axis from the larger one to get the focal distance, and eccentricity $e=c/a$ is always strictly between $0$ and $1$ . For a hyperbola, $c$ is the largest of the three lengths, because the foci lie outside the vertices, farther from the centre than the curve itself, so $c^2=a^2+b^2$ , hence $b^2=c^2-a^2$ : the direction of the subtraction is reversed from the ellipse. Students who have just finished the ellipse section instinctively write $b^2=a^2-c^2$ for a hyperbola too, which forces $b^2$ negative, since $c\gt a$ always in a hyperbola, and derails the entire problem. The fix is one sentence: in an ellipse the constant $2a$ beats the focal separation $2c$ , so $a\gt c$ and $e\lt 1$ , while in a hyperbola the focal separation beats the constant $2a$ , so $c\gt a$ and $e\gt 1$ ; always identify which of $a$ , $b$ , $c$ is "the odd one out," obtained by subtraction, before writing any formula.

You should be able to…

State the point-set definition of a circle, parabola, ellipse and hyperbola, and explain how each follows from slicing a cone at a particular angle.
Derive the standard equation of a parabola, ellipse or hyperbola from its focus-distance definition, including checking that the converse holds.
Identify, from a standard equation, the centre, foci, vertices, axes, eccentricity and latus rectum of any of the four conics.
Correctly apply $a^2=b^2+c^2$ for an ellipse and $b^2=c^2-a^2$ for a hyperbola without mixing up the direction of the subtraction.
Determine the orientation of a parabola, ellipse or hyperbola, which axis it opens along or lies on, from the signs and relative sizes in its equation.
Convert a circle's general equation $x^2+y^2+2gx+2fy+c=0$ into centre-radius form by completing the square.
Compute the latus rectum of any conic and know when to use $4a$ versus $2b^2/a$ .
Set up and solve for an unknown conic's equation given geometric data such as foci, vertices, eccentricity or latus rectum length, discarding extraneous roots.
Explain the unifying role of eccentricity across all four conics, including the circle and the parabola as limiting and boundary cases.

Ask the tutor: Want to work through converting a messy general-form conic equation into standard form step by step, or practice a JEE-style problem that mixes two conics in one question?

Why is the rule for the major axis of an ellipse based on the size of the denominator and not on whether x squared or y squared appears first?

Because the standard forms are written so that $a$ is always the semi-major axis and $b$ the semi-minor axis, and $a$ is always the larger of the two. If you see $\frac{x^2}{25}+\frac{y^2}{9}=1$ , the larger denominator, $25$ , sits under $x^2$ , so $a^2=25$ and the major axis is along the x-axis; if instead you saw $\frac{x^2}{9}+\frac{y^2}{25}=1$ , the larger denominator now sits under $y^2$ , so $a^2=25$ and the major axis is along the y-axis, even though the equation looks like the same shape at a glance. The position of the letters $x$ and $y$ tells you nothing about which axis is major; only comparing the two denominators does. This is exactly the observation NCERT itself flags, and examiners exploit it deliberately.

Why does an ellipse always have eccentricity less than 1 while a hyperbola always has eccentricity greater than 1?

Eccentricity is defined the same way for both, $e=c/a$ , but what $c$ and $a$ mean relative to each other differs. In an ellipse, the sum of the two focal distances is fixed at $2a$ , and by the triangle inequality that sum must exceed the distance $2c$ between the foci, so $a\gt c$ and $e=c/a\lt 1$ . In a hyperbola, the difference of the two focal distances is fixed at $2a$ , and the geometry works the opposite way: the foci sit outside the curve, farther from the centre than any vertex, forcing $c\gt a$ and hence $e=c/a\gt 1$ . A parabola sits exactly at the boundary between these two behaviours, with $e=1$ by definition.

Why does the latus rectum formula change from 4a for a parabola to 2b squared over a for an ellipse and a hyperbola?

The parabola has only one shape parameter, $a$ , so its latus rectum has to be expressed purely in terms of $a$ , and the derivation, setting $x=a$ in $y^2=4ax$ , gives $y^2=4a^2$ , $y=\pm2a$ , a full chord length of $4a$ . The ellipse and hyperbola each have two independent shape parameters, $a$ and $b$ , and substituting $x=c$ into their equations brings both into the answer; working through the algebra, as this chapter does for the ellipse, produces $l=b^2/a$ for the half-chord, so the full latus rectum is $2b^2/a$ . The two formulas are not really different rules so much as the same substitution method applied to equations with different numbers of parameters.

Is a circle really a conic section, since it does not seem to have a focus and a directrix like the other three?

Yes. A circle is the special case obtained when the cutting plane is exactly perpendicular to the cone's axis, $\beta=90^\circ$ , and algebraically it is the limiting case of an ellipse as the eccentricity shrinks to zero, at which point the two foci merge into a single centre and $a=b=r$ . Because the directrix of an ellipse recedes to infinity as $e\to0$ , a circle has no finite directrix in the usual focus-directrix sense, which is why this chapter introduces the circle with the simpler fixed-distance-from-a-centre definition rather than forcing focus-directrix language onto it. It is still, geometrically and historically, the first and simplest conic section.

How do I quickly tell, just from looking at an equation, which of the four conics it represents?

Check the powers and signs first. If both variables are squared with the same sign and equal coefficients, such as $x^2+y^2+\ldots=0$ , it is a circle. If both variables are squared with the same sign but different coefficients, such as $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ , it is an ellipse. If both variables are squared but with opposite signs, such as $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ , it is a hyperbola. If only one variable is squared and the other appears to the first power, such as $y^2=4ax$ , it is a parabola. This four-way check, done in under ten seconds, should be the very first thing you do on any conic-section problem, before touching any formula.

Conic Sections

What this chapter covers

The core ideas

Conic sections as slices of a cone

The circle

The parabola

The ellipse

The hyperbola

Eccentricity: one number that classifies every conic

Worked examples

Board level - centre and radius of a circle from the general equation

Board level - foci, vertices, axes, eccentricity and latus rectum of an ellipse

Competitive level (JEE / NEET) - hyperbola from foci and latus rectum, with an extraneous root

Common traps

You should be able to…

Before this

Where this leads