Figure 6.3. A Conic Section

Let $S(\alpha ,\beta )$ be the focus and $QN$ the directrix whose equation is $Ax+By+C=0$.

Let $P(x,y)$ be any point on the conic section. From $P$ draw $PN\perp QN$. Let $e$ be the eccentricity, then by definition $\frac{PS}{PN}=e\Rightarrow P{S}^2=e^{2}P{N}^2$.

$P{S}^2=(x-\alpha {)}^2+(y-\beta {)}^2$ and $P{N}^2=\frac{(Ax+By+C{)}^2}{A^2+B^2}$

$\Rightarrow (x-\alpha {)}^2+(y-\beta {)}^2=\frac{e^2(Ax+By+C{)}^2}{A^2+B^2}$, which on simplification will have the form $x^2+y^2+2gx+2fy+c=0$, where $a,b,c,f,g,h$ are constants.

Corollary: Let the focus $S$ lie on the directrix $QN$ and $S$ be taken as the origin and $QN$ be taken as the $y$-axis. Let $P(x,y)$ be any point on the conic then

$\frac{PS}{PN}=e\Rightarrow \frac{\sqrt{x^2+y^2}}{|x|}=e\Rightarrow (e^2-1)x^2=y^2$, which is a pair of straight lines through the focus or the origin.

Corollary: The general equation of the second degree $a{x}^2+2y^2+2hxy+2gx+2fy+c=0$ represents a parabols if $h^2=ab$ and $\mathrm{\Delta }\ne 0$, where $\mathrm{\Delta }=abc+2fgh-a{f}^2-b{g}^2-c{h}^2$ or alternatively a second degree equation represents a parabola if second degree terms form a complete square and $\mathrm{\Delta }neq0$.

The second degree equation will represent an ellipse if $h^2<ab$. We will see that a circle is a special case of an ellipse.

The second degree equation will represent a hyperbola if $h^2>ab$ and $\mathrm{\Delta }\ne 0$ and a pair of straight lines if $\mathrm{\Delta }=0$.

Consider a straight line given by the equation $y=mx+c$ and a parabola given by the equation $y^2=4ax$.

Solving the equations gives us $(mx+c{)}^2=4ax\Rightarrow m^2{x}^2+2x(mc-2a)+c^2=0$.

The above equation, which is a quadratic equation in $x$ will have two roots. If the roots are real i.e. $(mc-2a{)}^2-m^2{c}^2>0$, then the will cut the parabola at two different points. If $(mc-2a{)}^2-m^2{c}^2=0$, then there will be a repeated root and the line will be a tangent to the parabola and if $(mc-2a{)}^2-m^2{c}^2<0$, then roots will be imaginary and the line will not inttersect with the parabola.

We see that $(mc-2a{)}^2-m^2{c}^2=0$ is the condition for tangency. Thus, $$c=\frac{a}{m}$$(6.2)

Thus, any line of the form $y=mx+a/m$ will be a tangent to the parabola.

Let the equation of parabola be $y^2=4ax$. Let $P=(x_1,y_1)$ and $Q(x_2,y_2)$ be any two points on the parabola. The equation of $PQ$ will be given by

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\dots$ (1)

Since both the points lie on the parabola, therefore $y_1^2=4a{x}_1$ and $y_2^2=4a{x}_2$

Subtracting these two equations yields $(y_2+y_1)(y_2-y_1)=4a(x_2-x_1)\Rightarrow \frac{y_2-y_1}{x_2-x_1}(x-x_1)=\frac{4a}{y_2+y_1}$

Figure 6.5. A Tangent to a Parabola

Putting this in (1) gives us $y-y_1=\frac{4a}{x_2+x_1}(x-x_1)$

Now, when $Q\to P,x_2\to x_1$ and $y_2\to y_1$ so the above equation becomes

$y-y_1=\frac{4a}{2y_1}(x-x_1)\Rightarrow y{y}_1=2ax+y_1^2-2a{x}_1$

Substituting $y_1^2=4a{x}_1$ we have $$y{y}_1=2a(x+x_1)$$(6.3)

which is the equation of the required tangent.

We have obtained the equation of the tangent to a parabola $y^2=4ax$ at $(x_1,y_1)$ as $y{y}_1=2a(x+x_1)\dots$ (1), The slope of tangent is $\frac{2a}{y1}$.

The normal will be perpendicular to the tangent, and hence, its slope will be $-\frac{y_1}{2a}$.

Thus equation of normal would be $$y-y_1=-\frac{y_1}{2a}(x-x_1)$$(6.4)

The normal to the parabola $y^2=4ax$ at $(x_1,y_1)$ is $y-y_1=-\frac{y_1}{2a}(x-x_1)$.

Let $-\frac{y_1}{2a}=m\Rightarrow y_1=2am$. Since $y_1^2=4a{x}_1\Rightarrow 4a^2{m}^2=4a{x}_1\Rightarrow x_1=a{m}^2$

$\Rightarrow y-(-2am)=m(x-a{m}^2)\Rightarrow y=mx-2am-a{m}^3$, which is the equation of the normal in terms of its slope.

Notes: The equation of the normal obtained is $y=mx-2am-a{m}^3$. This normal passes through $(x_1,y_1)$. Thus, $a{m}^3+(2a-x_1)+m+y_1=0$.

This is a cubic equation in $m$, which will have three roots of which at least one will be real because complex roots occur in pair.

Any three points on the parabola, normals at which pass through a common point, are called co-normal points.

The point $P$ whose coordinates are $x=a{t}^2,y=2at$ lies on the parabola $y^2=4ax$, which can be trivially verified by substituting in $y^2=4ax$.

This representation in which $x$ and $y$ are represented using a single parameter $t$ is called the parametric equation of the parabola.

The point $(a{t}^2,2at)$ is also referred to as the point '$t$'.

Consider two points $(a{t}_1^2,2a{t}_1)$ and $(a{t}_2^2,2a{t}_2)$ on the parabola forming a chord.

The equation will be given by $y-2a{t}_1=\frac{2a{t}_2-2a{t}_1}{a{t}_2^2-a{t}_1^2}(x-a{t}_1^2)$

$$y(t_1+t_2)=2x+2a{t}_1{t}_2$$(6.5)

is the equation of the the chord.

If this line passes through the focus i.e. $(a,0)$ then $2a+2a{t}_1{t}_2=0\Rightarrow t_1{t}_2=-1$. Thus, if one point is $t$ then the condition for the chord to pass through the focus requires the other point to be $-\frac{1}{t}$.

The tangent to the parabola $y^2=4ax$ at the point $(a{t}^2,2at)$ is given by

$y(2at)=2a(x+a{t}^2)\Rightarrow ty=x+a{t}^2$.

The normal to the parabola $y^2=4ax$ at the point $(a{t}^2,2at)$ is given by

$y-2at=-t(x-a{t}^2)\Rightarrow y+tx=2at+a{t}^3$.

We see that two tangents at $t_1$ and $t_2$ have $(a{t}_1{t}_2,a(t_1+t_2))$ as their point of intersection and the normals at the same points will have $(2a+a(t_1^2+t_2^2+t_1{t}_2),-a{t}_1{t}_2(t_1+t_2))$ as their point of intersection.

Figure 6.6. Chord of a parabola

Equation of the parabola is $y^2=4ax$. Let $AB$ be a chord of the parabola whose middle point is $P(x_1,y_1)$.

Equation of the chord is $y-y_1=m(x-x_1)$, where $m$ is the slope of $AB$.

Let $A=(x_2,y_2)$ and $B=(x_3,y_3)\Rightarrow y_2^2=4a{x}_2$ and $y_3^2=4a{x}_3$

$\therefore y_2^2-y_3^2=4a(x_2-x_3)\Rightarrow \frac{y_2-y_3}{x_2-x_3}=\frac{4a}{y_2+y_3}$

But $P$ is mid-point of $AB$, so we can write $y_2+y_3=2y_1$

$\Rightarrow \frac{y_2-y_3}{x_2-x_3}=\frac{2a}{y_1}=m$

Thus, equation of the chord is $y-y_1=\frac{2a}{y_1}(x-x_1)\Rightarrow y{y}_1-2a(x+x_1)=y_1^2-4a{x}_1$.

Figure 6.7. Position of a point w.r.t. a parabola

Let $P(x_1,y_1)$ be a given point. Let $PL\perp OX$ and let $PL$ intersect the parabole at $Q(x_2,y_2)$. Since $Q(x_2,y_2)$ lies on the parabola, therefore $y_2^2=4a{x}_2$.

Point $P$ will lie outside, on or inside the parabola $<=>PL>,=,\text{or}<QL$

$|y_1|>,=\text{or}<|y_2|=>y_1^2-4a{x}_1>,=,\text{or}<0$.

Let the parabola be $y^2=4ax$ and $P(\alpha ,\beta )$ be a given point.

The equation of a tangent to the parabola is $y=mx+\frac{a}{m}$

$\Rightarrow m^2\alpha -\beta m+a=0.D>0\Rightarrow {\beta }^2-4a\alpha >0$ i.e. if the point lies outside the parabola then $m$ will be real and the equation will have two real roots, which gives us two tangents to the parabola from the point $P$.

Figure 6.8. Pair of tangents from a point w.r.t. to a parabola

Let the parabola be $y^2=4ax$. Let $P(x_1,y_1)$ be a point outside the parabola. Let a chord of the parabola through the point $P$ cut the parabola at $R$ and let $Q(\alpha ,\beta )$ be any arbitrary point on the line $PR$. Let $R$ divide $PQ$ in the ratio $\lambda :1$, then

$R=(\frac{\lambda \alpha +x_1}{\lambda +1},\frac{\lambda \beta +y_1}{\lambda +1})$

Since $R$ lies on the parabola, therefore ${\left(\frac{\lambda \beta +y_1}{\lambda +1}\right)}^2=4a\frac{\lambda \alpha +x_1}{\lambda +1}$

$\Rightarrow ({\beta }^2-4a\alpha ){\lambda }^2+2[\beta y_1-2a(\alpha +x_1)]\lambda +(y_1^2-4a{x}_1)=0\dots$ (1)

Line $PQ$ will become a tangent to the parabola if the roots of the above equation (1) are equal

$\Rightarrow 4[\beta y_1-2a(\alpha +x_1){]}^2=4({\beta }^2-4a\alpha )(y_1^2-4a{x}_1)$

Hence, locus of $(\alpha ,\beta )$ i.e. equation of pair of tangents from $P$ is $[y{y}_1-2a(x+x_1){]}^2=(y^2-4ax)(y_1^2-4a{x}_1)$

$\Rightarrow T^2=S{S}_1$, where $T,S$ and $S_1$ have usual meanings.

Let the parabola be $y^2=4ax\dots$ (1) Let $P(\alpha ,\beta )$ be a point outside the parabola. Let $PA$ and $PB$ be the two tangents from $P$ to the parabola (1). Let $PA$ and $PB$ be the two tangents from $P$ to the parabola (1).

Figure 6.9. Chord of contact from a point w.r.t. to a parabola

Let $A=(x_1,y_1)$ and $B=(x_2,y_2)$. Equation of the tangent $PA$ is $y{y}_1=2a(x+x_1)$ and that of $PB$ is $y{y}_2=2a(x+x_2)$.

Since both the lines pass through $P$, therefore $\beta y_1=2a(\alpha +x_1)$ and $\beta y_2=2a(\alpha +x_2)$.

Now we consider the equation $y\beta =2a(x+\alpha )$. We see that this line passes through both $A$ and $B$. Therefore, it is the equation of chord of contact $AB$ of point $P$ w.r.t. the parabola.

Figure 6.10. Pole and polar of a parabola

Let $P$ be a given point. Let a line through $P$ intersect parabola at two points $A$ and $B$. Let the tangents at $A$ and $B$ meet at $Q$. The locus of $Q$ is a straight line called the polar of point $P$.

If the line $LM$ is the polar of a point $P$ w.r.t. a parabola, then the point $P$ is called the pole of the line $LM$ w.r.t. the parabola.

Let the parabola be $y^2=4ax\dots$ (1) Let $P(x_1,y_1)$ be the given point. Let a line though $P$ cut parabola in two points $A$ and $B$.

Let $Q(\alpha ,\beta )$ be the point of intersection of the tangents to the parabola at $A$ and $B$.

Clearly, $AB$ is the chord of contact of point $Q(\alpha ,\beta )$ w.r.t. the parabola. Thereforem, equation of $AB$ will be

$y\beta =2a(x+\alpha )\dots$ (2)

Since this line passes through $P(x_1,y_1)$, therefore $y_1\beta =2a(x_1+\alpha )$.

Hence, locus of $Q(\alpha ,\beta )$ i.e. equation of polar of point $P$ w.r.t. the parabola is

$$y{y}_1=2a(x+x_1)$$(6.6)

According to the above definition of polar, $AB$ will not be a part of the locus because no two tangents to the parabola can intersect on the line segment $AB$.

So the correct definition is here. Let $P(x_1,y_1)$ be any point. The line whose equation is that of the tangent to $P$ to parabola whether $P$ lies on the curve or not is called the polar of point $P$ w.r.t. the parabola. According to this definition equation of polar of point $P(x_1,y_1)$ w.r.t. the parabola is $T=0$ or $y{y}_1=2a(x+x_1)$.

Diameter of a conic is the locus of mid-points of a series of its parallel chords.

Figure 6.11. Diameter a parabola

Let the parabola be $y^2=4ax\dots$ (1)

Let $AB$ be one of the chords of a series of parallel chords having slope $m$. Let $P(\alpha ,\beta )$ be the mid-point of chord $AB$, then equation of $AB$ will be $T=S_1$.

$\Rightarrow y\beta -2a(x+\alpha )={\beta }^2-4a\alpha \dots$ (2)

Slope of this line is $frac2a\beta$. But slope of the line $AB$ is $m$, therefore $\frac{2a}{\beta }=m\Rightarrow beta=\frac{2a}{m}$.

Hence, locus of $P(\alpha ,\beta )$ i.e. equation of diameter is $$y=\frac{2a}{m}$$(6.7)

Clearly, this line is parallel to the axis of the parabola.

The tangent at any point $P$ on a parabola bisects the angle between the focal chord through $P$ and the perpendicular from $P$ on the directrix.

Let the parabola be $y^2=4ax\dots$ (1)

Let $P(a{t}^2,2at)$ be a point on the parabola and let $PM\perp \text{ directrix}$.

Figure 6.12. 

Equation of the tangent to parabola at $P$ is $yt=x+a{t}^2$. Slope of this tangent $PT$ is $\frac{1}{t}\Rightarrow \tan \theta =\frac{1}{t}$

Slope of $PS=\frac{2at}{a(t^2-1)}=\frac{\frac{2}{t}}{1-\frac{1}{t^2}}=\frac{2\tan \theta }{1-{\tan }^2\theta }=\tan 2\theta$.

$\Rightarrow \mathrm{\angle }PSX=2\theta \Rightarrow \mathrm{\angle }TPS=\mathrm{\angle }PSX-\mathrm{\angle }PTS=2\theta -\theta =\theta$.

Hence, $\mathrm{\angle }SPT=\mathrm{\angle }MPT=\theta$.

Therefore $PT$ bisects $\mathrm{\angle }MPS$.

The other property we know from physics is that rays coming from infinity are reflected by parabolas in such a manner that they pass through focus. We will prove that here.

Let $PN$ be normal to $PT$. In $\mathrm{△}PTN,\mathrm{\angle }TPN={90}^{\circ }$ and $\mathrm{\angle }PTN=\theta$

Since $PR\parallel TN,\therefore \mathrm{\angle }RPN=\mathrm{\angle }TNP={90}^{\circ }-\theta$

Also, $\mathrm{\angle }SPN={180}^{\circ }-\mathrm{\angle }PSN-\mathrm{\angle }SNP={180}^{\circ }-2\theta -({90}^{\circ }-\theta )={90}^{\circ }-\theta$

Figure 6.13. 

$\therefore \mathrm{\angle }SPN=\mathrm{\angle }RPN$.

Thus, a ray parallel to the axis of parabola after reflection from parabola passes through the focus.

Let $S$ be the focus and $M{M}^{\prime }$ be the directrix of the ellipse. Draw $SH\perp M{M}^{\prime }$, wed divide $SH$ internally and externally in the ratio $e:1(e<1)$ at points $A$ and $A^{\prime }$ respectively, then

Figure 6.15. Equation of an ellipse

$\frac{AS}{AH}=e\dots$ (1) and $\frac{A^{\prime }S}{A^{\prime }H}=e\dots$ (2)

Clearly, $A$ and $A^{\prime }$ will lie on the ellipse. Let $A{A}^{\prime }=2a$. Let $O$ be the mid-point of $A{A}^{\prime }$. We take $O$ as the origin and $O{A}^{\prime }$ and $OY$ as the $x$ and $y$ axes respectively.

From (1), $AS=eAH$ and from (2) $A^{\prime }S=e{A}^{\prime }H$.

$AS+A^{\prime }S=e(AH+A^{\prime }H)\Rightarrow 2a=(AH+AH+A{A}^{\prime })=e(2Ah+2a)\therefore AH=\frac{a}{e}-e$.

From (1) $AS=eAH=a-ae\therefore OS=OA-AS=ae\therefore S=(-ae,0)$.

and $OH=OA+AH=\frac{a}{e}$.

Equation of $M{M}^{\prime }$ is $x=-\frac{a}{e}\Rightarrow ex+a=0$.

Let $P(\alpha ,\beta )$ be a point on ellipse. Draw $PN\perp M{M}^{\prime }$

Now $\frac{PS}{PN}=e\Rightarrow P{S}^2=e^{2}P{N}^2$

$\Rightarrow (\alpha +ae{)}^2+bet{a}^2=e^2{\left(\frac{|ealpha+a|}{e}\right)}^2$

$\Rightarrow \frac{{\alpha }^2}{a^2}+\frac{{\beta }^2}{a^2(1-e^2)}=1$.

Let $b^2=a^2(1-e^2)$

Then the locus of $P$ is the equation of the ellipse, which is

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$(6.8)

It is clear that if $(ae,0)$ is the focus, and $ex-a=0$ is the directrix even then the ellipse will have the same equation.

Thus, the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ has two foci and two directrices.

A circle is a special limiting case of an ellipse whose major axis is $2a$, both foci approach the center and eccentricity approaches $0$.

A typical ellipse is given below:

Figure 6.16. An ellipse

As you can see an ellipse cuts $x$-axis at $(-a,0)$ and $(a,0)$, cuts $y$-axis at $(-b,0)$ and $(b,0)$. Equation of ellipse does not change when $x$ or $y$ are changed to $-x$ and $-y$, hence, an ellipse is symmetrical about $x$ and $y$ axes.

We have $y=\pm \frac{b}{a}\sqrt{a^2-x^2}\Rightarrow a^2-x^2\ge 0\Rightarrow -a\le x\le a$.

Similarly, $x=\pm \frac{a}{b}\sqrt{b^2-y^2}\Rightarrow b^2-x^2\ge 0\Rightarrow -b\le y\le b$.

We see that ellipse is bounded between $-a\le x\le a$ and $-b\le y\le b$ so an eclipse is a closed curve.

$\frac{dy}{dx}=-\frac{b^{2}x}{b^{2}y}$. At $(-a,0)$ and $(a,0)$, the slope is not defined, and hence, tangent at $(-a,0)$ and $(a,0)$ will be perpendicular to $x$-axis.