Introduction to Conic Sections

Conic sections can be generated when one slanted line revolves around an intersecting a vertical line . We get two cones joined at the intersection and expading indefinitely upward and downward.

When a plane intersects the created double cone, we have a conic section at the intersection.

Depending on how the plane is intersecting the conic section, we get different types of conic sections.

The conic sections may be : parabola, ellipse, hyperbola, circle etc…

The following is a summary to enable us to solve problems involving the conic sections.

The Parabola:

A parabola can be defined as the collection of all points $P$ in the plane that are the same distance $d$ from a fixed point $F$ as they are from another fixed line $D$ .

The point $F$ is called the $Focus$

The line $D$ is the $Directrix$ .

$\fbox{d(F,P)=d(P,D)}$

The line through the focus $F$ and perpendicular to the directrix $D$ is the $axis \; of \; symmetry$ .

The point of intersection of the parabola and its axis of symmetry is the $VERTEX$ , noted $V$ .

If we take a VERTEX at $(0,0)$ , with FOCUS at $(a,0)$ while $a>0$ , by definition the DIRECTRIX will be the line $x=-a$ .

Using the definition above and the distance formula:

$\sqrt{(x-a)^{2}+(y-0)^{2}}=|x+a|$

Solving we get:

Equation of the parabola:

$y^{2}=4ax$

The LATUS RECTUM here has a length of $4a$ and is defined as the line segment joining $(a,2a)$ and $(a,-2a)$

We get the following cases: for a VERTEX at (h,k):

-When for vertex $(h,k)$ we have a focus $(h+a,k)$ , the DIRECTRIX will be the line $x=h-a$ and the equation is $(y-k)^{2}=4a(x-h)$ Symmetry axis is parrallel to $x-axis$ , opens right.

-When for vertex $(h,k)$ we have a focus $(h-a,k)$ , the DIRECTRIX will be the line $x=h+a$ and the equation is $(y-k)^{2}=-4a(x-h)$ Symmetry axis is parrallel to $x-axis$ , opens left.

-When for vertex $(h,k)$ we have a focus $(h,k+a)$ , the DIRECTRIX will be the line $y=k-a$ and the equation is $(x-h)^{2}=4a(y-k)$ Symmetry axis is parrallel to $y-axis$ , opens up.

-When for vertex $(h,k)$ we have a focus $(h,k-a)$ , the DIRECTRIX will be the line $y=k+a$ and the equation is $(x-h)^{2}=-4a(y-k)$ Symmetry axis is parrallel to $y-axis$ , opens down.

We get the following cases: for a VERTEX at (0,0):

-When for vertex $0,0)$ we have a focus $(a,0)$ , the DIRECTRIX will be the line $x=-a$ and the equation is $y^{2}=4ax$ Symmetry axis is $x-axis$ , opens right.

-When for vertex $0,0)$ we have a focus $(-a,0)$ , the DIRECTRIX will be the line $x=a$ and the equation is $y^{2}=-4ax$ Symmetry axis is $x-axis$ , opens left.

-When for vertex $0,0)$ we have a focus $(0,a)$ , the DIRECTRIX will be the line $y=-a$ and the equation is $x^{2}=4ay$ Symmetry axis is $y-axis$ , opens up.

-When for vertex $0,0)$ we have a focus $(0,-a)$ , the DIRECTRIX will be the line $y=a$ and the equation is $x^{2}=-4ay$ Symmetry axis is $y-axis$ , opens down.

The Hyperbola:

A hyperbola can be defined as the collection of all points $P$ in the plane, haveing a difference of distances from two points, $FOCI$ as constant.

The line containing the Foci is the $TRANSVERSE \; AXIS$ .

The $CENTER$ of the hyperbola is the midpoint of the segment joining the $FOCI$ .

The perpendicular of the transverse axis trhough the center is called the $CONJUGATE \; AXIS$ .

The two separate curves of the hyperbola are the $BRANCHES$

The points of intersection of the hyperbola and the transverse axis are the $VERTICES$ of the hyperbola.

If $F_{1}(-c,0)$ , $F_{2}(c,0)$ with $c$ the distance between a focus and the center.

For any given point $H(x,y)$ , we can have:

$d(H,F_{1})-d(H,F_{2})=\pm 2a$

Using the distance formula:

In this equation we have $2a<2c \Rightarrow (a<c)$

$\sqrt{((-c)-x)^{2}+(0-y)^{2}}-\sqrt{(c-x)^{2}+(0-y)^{2}}=\pm2a$

$\sqrt{c^{2}+2cx+x^{2}+y^{2}}=\pm 2a+\sqrt{c^{2}-2cx+x^{2}+y^{2}}$

Let’s square both sides:

$c^{2}+2cx+x^{2}+y^{2}=4a^{2} \pm4a\sqrt{c^{2}-2cx+x^{2}+y^{2}}+ {c^{2}-2cx+x^{2}+y^{2}}$

We simplify:

$4cx=4a^{2}\pm 4a\sqrt{c^{2}-2cx+x^{2}+y^{2}}$

Simplifying by 4

$cx=a^{2}\pm a\sqrt{c^{2}-2cx+x^{2}+y^{2}}$

$cx-a^{2}=\pm a\sqrt{c^{2}-2cx+x^{2}+y^{2}}$

Let’s square both sides again:

$c^{2}x^{2}-2ca^{2}x+a^{4}=a^{2}(c^{2}-2cx+x^{2}+y^{2})$

$c^{2}x^{2}-2ca^{2}x+a^{4}=a^{2}c^{2}-2ca^{2}x+a^{2}x^{2}+a^{2}y^{2})$

$c^{2}x^{2}+a^{4}=a^{2}c^{2}+a^{2}x^{2}+a^{2}y^{2}$

$c^{2}x^{2}+a^{4}-a^{2}x^{2}=a^{2}c^{2}+a^{2}y^{2}$

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

Let: $b^{2}=c^{2}-a^{2}$

We get:

$x^{2}b^{2}-a^{2}y^{2}=a^{2}b^{2}$

Now dividing both sides by $a^{2}b^{2}$

We have the hyperbola formula:

$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$

The transverse axis here is the $x-axis$

Asymptotes:

We’ll see in future chapters how to calculate the limits.

However, if we take:

$y=\pm \frac{bx}{a}\sqrt{1-\frac{a^{2}}{x^{2}}}$

When $x$ approaches infinity (left or right side), the value $\frac{a^{2}}{x^{2}}$ approaches $0$

we get:

$y=-\frac{b}{a}x$ and $y=-\frac{b}{a}x$

These are the two oblique asymptotes:

$y=\frac{b}{a}x$ and $y=-\frac{b}{a}x$

For the hyperbola of Center at $(0,0)$ ; The transverse Axis along the $y-Axis$

Foci are at $(0,-c)$ and $(0,c0$ ,

and vertices at $(0,-a)$ and $(0,a)$

$\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$

Asymptotes:

We’ll see in future chapters how to calculate the limits.

However, if we take:

$y=\pm \frac{ax}{b}sqrt{1-\frac{b^{2}}{x^{2}}}$

When $x$ approaches infinity (left or right side), the value $\frac{b^{2}}{x^{2}}$ approaches $0$

we get:

$y=-\frac{a}{b}x$ and $y=-\frac{a}{b}x$

These are the two oblique asymptotes:

$y=\frac{a}{b}x$

and

$y=-\frac{a}{b}x$

Hyperbola Equations:

Transverse parallel to the $x-axis$

Center $(h,k)$

Foci: $(h \pm c,k)$

Vertices: $(h \pm c,k)$

Equation:

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

$b^{2}=c^{2}-a^{2}$

Asymptotes:

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

Transverse parallel to the $y-axis$

Center $(h,k)$

Foci: $(h ,k \pm c)$

Vertices: $(h ,k \pm c)$

Equation:

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

$b^{2}=c^{2}-a^{2}$

Asymptotes:

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

The ELLIPSE:

The Ellipse can be defined as the collection of all points $P$ in the plane, having a sum of distances from two points, $FOCI$ as constant.

The line containing the Foci is the $MAJOR \; AXIS$ .

The $CENTER$ of the ellipse is the midpoint of the segment joining the $FOCI$ .

The perpendicular of the major axis trhough the center is called the $MINOR \; AXIS$ .

The points of intersection of the ellipse and themajor axis are the $VERTICES$ of the ellipse.

If $F_{1}=(-c,0)$ and $F_{2}=(c,0)$ , $2a$ is the constant distance: Point $H(x,y)$

$d(F_{1},H)+d(F_{2},H)=2a$

Using the same technic we used for the Hyperbola:

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

$b^{2}=a^{2}-c^{2}$ where $a>b>0$

The major axis here is along the $x-axis$

For the Ellipse of Center at $(0,0)$ ; The major Axis along the $y-Axis$

Foci are at $(0,-c)$ and $(0,c0$ ,

and vertices at $(0,-a)$ and $(0,a)$

$b^{2}=a^{2}-c^{2}$ where $a>b>0$

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

Ellipse Equations:

Major axis parallel to the $x-axis$

Center $(h,k)$

Foci: $(h \pm c,k)$

Vertices: $(h \pm a,k)$

Equation:

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

$b^{2}=a^{2}-c^{2}$

Major axis parallel to the $y-axis$

Center $(h,k)$

Foci: $(h ,k \pm c)$

Vertices: $(h ,k \pm a)$

Equation:

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

$b^{2}=a^{2}-c^{2}$

The Circles

A circle is the set of all points in a plane that are equidistant from a given point in the plane, $THE \; CENTER$

The distance from any point to the center is $RADIUS$ .

For a circle of center $(h,k)$ .

For a point $P(x,y)$

The radius $r$ can be found using the distance formula;

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

We square both sides:

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

The Equation of the circle of center $(h,k)$

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

The tangent is perpendicular to the radius at the tangency point.

For more details see our course on CIRCLE ESSENTIALS.

Conics, general equation

Conics can be identified using the general equation when we do not have a degenerating case:

$Ax^{2}+Cy^{2}+Dx+Ey+F=0$

$A$ and $C$ cannot both equal $0$

When $AC=0$ , we have a PARABOLA

When $AC>0$ we have an ELLIPSE (or CIRCLE)

When $AC<0$ , it is a hyperbola

We can also use the following rule to identify a conic:

When we do not have a degenerating case:

$Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$

When $B^{2}-4AC=0$ , we have a PARABOLA

When $B^{2}-4AC<0$ we have an ELLIPSE (or CIRCLE)

When $B^{2}-4AC>0$ , it is a hyperbola

Conics in polar Equations

In this passage, we are going to express the conics dicussed above, using their polar notations.

For a given Conic, we will call the line $D$ the $directrix$ , $F$ a given $Focus$ .

In our new notation, let’s introduce $e$ as a fixed positive number, the $eccentricity$ .

A $conic$ is then the set of points $H$ such the ratio of the distance $FH$ to the distance $DH$ is $e$ .

$e=\frac{d(F,H)}{d(D,H)}$

– For the parabola, we can note that $e=1$

-For the ellipse we have $e<1$

-For the hyperbola $e>1$

If $c$ is the distance from the center to the focus and $a$ is the distance from the center to the vertex, for ellipse and hyperbola:

$e=\frac{c}{a}$

Each point $H(x,y)$ can be expressed in polar coordinates as $H(r, \theta)$

$e=\frac{d(F,H)}{d(D,H)} \Rightarrow d(F,H)=e.d(D,H)$

If we drop a perpendicular from $H$ to the polar axis with $Q$ as intersection, we can write:

$d(D,H)=p+d(O,Q)=p+r\cos \theta$

But we can see that $d(F,H)=d(O,H)=r$

We can plug in:

$d(F,H)=e.d(D,H)$

$r=e.(r+ \cos \theta)$

Solving for $r$ we get:

The polar equation of a Conic:

$r=\frac{ep}{1-e \cos \theta}$

With $e$ the eccentricity of the conic.

If the $directrix$ is perpendicular to the polar-axis at a distance $p$ on the right of the pole:

we get:

$r=\frac{ep}{1+e \cos \theta}$

SUMMARY

1. Directrix perpendicular to the polar axis at $p$ units to the left of the pole:

$\mathbf{\displaystyle r=\frac{ep}{1-e \cos \theta}}$

2. Directrix perpendicular to the polar axis at $p$ units to the right of the pole:

$\mathbf{\displaystyle r=\frac{ep}{1+e \cos \theta}}$

3. Directrix parallel to the polar axis at $p$ units above the pole:

$\mathbf{\displaystyle r=\frac{ep}{1+e \sin \theta}}$

4. Directrix parallel to the polar axis at $p$ units below the pole:

$\mathbf{\displaystyle r=\frac{ep}{1-e \sin \theta}}$

Please note:

-For $e=1$ , the conic is a parabola, the axis of symmetry is perpendicular to the $directrix$

-For $e<1$ , the conic is an ellipse. The major axis is perpendicular to the $directrix$

-For $e>1$ , the conic is a hyperbola. The transverse axis is perpendicular to the $directrix$

Converting from polar to rectangular and vice-versa

We use the following relationships:

$x=r\cos \theta$

$y=r\sin \theta$

$\tan \theta=\frac{y}{x}$

And of course:

$x^{2}+y^{2}=r^{2}$

Post Views: 278

Introduction to Conic Sections

Introduction to Conic Sections

The Parabola:

The Hyperbola:

Asymptotes:

Asymptotes:

Hyperbola Equations:

The ELLIPSE:

$d(F_{1},H)+d(F_{2},H)=2a$

Ellipse Equations:

The Circles

Conics, general equation

$Ax^{2}+Cy^{2}+Dx+Ey+F=0$

$Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$

Conics in polar Equations

$e=\frac{d(F,H)}{d(D,H)}$

$e=\frac{c}{a}$

The polar equation of a Conic:

$r=\frac{ep}{1-e \cos \theta}$

$r=\frac{ep}{1+e \cos \theta}$

Converting from polar to rectangular and vice-versa

Be the first to comment

Leave a Reply Cancel reply