conic sectionsgeometryproof-verification
independent from the directrix,
the eccentricity is defined as follows:
For a given ellipse:
the length of the semi-major axis = $a$
the length of the semi-minor = $b$
the distance between the foci = $2c$
the eccentricity is defined to be $\dfrac{c}{a}$
now the relation for eccenricity value in my textbook is $\sqrt{1- \dfrac{b^{2}}{a^{2}}}$
which I cannot prove.
The figure below shows a conic section (specifically an ellipse) with focus $F$ and directrix $D$. The conic is the intersection of the blue cone and the purple "cutting plane". The figure also shows the Dandelin sphere associated with $F$ and $D$. (Interestingly, the Wikipedia article mentions the focus-directrix property "can be proved" with Dandelin spheres, but doesn't give that proof!) Let's call the green "horizontal" plane containing the circle where the sphere meets the cone the Dandelin plane; call that circle the Dandelin circle.
Now, given $P$ on the conic, let $Q$ be the corresponding point on the Dandelin circle; that is, let $Q$ be the point where the segment joining $P$ to the cone apex meets the Dandelin plane. Let $R$ be the foot of the perpendicular dropped from $P$ to the Dandelin plane, and let $S$ be the foot of the perpendicular dropped from $P$ to the directrix.
(Original image credit, with description of Dandelin spheres.)
Since segments $\overline{PF}$ and $\overline{PQ}$ are both tangent to the Dandelin sphere, we must have $|\overline{PF}|=|\overline{PQ}|$. (This, by the way, is the primary magic of the Dandelin sphere.)
We can massage the focus-directrix ratio for $P$ thusly:
$$\frac{|\overline{PF}|}{|\overline{PS}|} = \frac{|\overline{PQ}|}{|\overline{PS}|}=\frac{|\overline{PR}|/(\sin\angle Q)}{|\overline{PS}|}=\frac{|\overline{PR}|}{|\overline{PS}|}\frac{1}{\sin\angle Q}=\frac{\sin \angle S}{\sin \angle Q} \tag{$\star$}$$
Clearly, $\angle S$ is constant as $P$ moves about the conic; it's the angle between the cutting plane and Dandelin plane. But $\angle Q$ is also constant: it's the ("exterior") angle that the surface (more precisely, a "generator" line) of the cone makes with the Dandelin plane. Therefore, the focus-directrix ratio is a constant.
To complete the answer to your question, all we have to do is prove that "(focal distance)-over-(major radius)" gives the same trigonometric ratio.
For now, we'll assume the conic is an ellipse (that is $\angle S$ is smaller than $\angle Q$).
Look at the figure "sideways", reducing all the elements to their intersections with the plane through the cone's axis, perpendicular to the directrix. I'll take $P$ to be the point on the conic closest to the directrix (which itself has projected into the point $S$), and $P^\prime$ the farthest point. ($Q$ and $Q^\prime$ are the corresponding points on the Dandelin plane, which has projected into a line.) Then $\overline{PP^\prime}$ is the major axis of the ellipse. The ellipse's focus, $F$, corresponds to the point where the incircle of $\triangle OPP^\prime$ meets $\overline{PP^\prime}$; the ellipse's center corresponds to $M$, the midpoint of $\overline{PP^\prime}$.
Now that we know where everything is, a couple more applications of the equal-tangent-segment property are all we need. With $a := |\overline{MP^\prime}|$ and $c := |\overline{MF}|$, we have
$$\frac{\sin\angle S}{\sin\angle Q} = \frac{|\overline{PT}|}{|\overline{PP^\prime}|} = \frac{|\overline{QT}| - |\overline{QP}|}{2a} = \frac{(a+c)-(a-c)}{2a}= \frac{2c}{2a}= \frac{c}{a} \tag{$\star\star$}$$
For the hyperbola, overlapping elements muddle the diagram a bit, but the argument is essentially the same (with a simple sign change):
$$\frac{\sin\angle S}{\sin\angle Q} = \frac{|\overline{PT}|}{|\overline{PP^\prime}|} = \frac{|\overline{QT}| + |\overline{QP}|}{2a} = \frac{(a+c)+(c-a)}{2a}= \frac{2c}{2a}= \frac{c}{a} \tag{$\star\star^\prime$}$$
There's no argument to make for the parabola, which has no "focal distance" or "major radius". However, one sees that, as the $\angle S$ nears $\angle Q$, the ratio of the lengths of these elements within an ellipse or hyperbola approaches $1$, as expected.
Thus, the "distance-to-focus-over-distance-to-directrix" ratio and the "focal-radius-over-major-radius" ratio (when defined) are the same constant that we happen to call the "eccentricity" of a conic. This discussion reveals the geometric meaning of that number. I suspect that most students these days had no idea that there is such meaning. Kudos to the teacher who assigned this problem as homework.
The equation $$ax^2+2hxy+by^2=c$$ represents a conic rotated about the axis (depending on $a,b,h,c$) with centre $(0,0)$.
You can always rotate the axis to remove the $xy$ term. This makes it easy to find the focus, length of major axis, etc.
To remove $xy$ term without shifting origin, lets rotate the axis by an angle $\theta$.
Let $X,Y$ represent the new coordinate system. Then, $$x=X\cos\theta-Y\sin\theta$$ $$y=X\sin\theta+Y\cos\theta$$ Substituting and letting coefficient of $XY=0$, you get $$\tan2\theta=\frac{2h}{a-b}$$
For example, lets find $e$ of the following equation (which represents an ellipse): $$4x^2+2y^2+2\sqrt{3}xy=1$$
Raotate the axis by $$\tan2\theta=\frac{2h}{a-b}=\sqrt{3}$$ $$\theta=\frac{\pi}{6}$$
Now, substitute $$x=X\frac{\sqrt{3}}{2}-Y\frac{1}{2}$$ $$y=Y\frac{\sqrt{3}}{2}+X\frac{1}{2}$$ You will get $$5X^2+Y^2=5$$ Now you can find eccentricity using the formula you specified. (Rotating the ellipse or hyperbola will not change the eccentricity and the length of major\minor axis.
This is the graph for the two equations:
(Black represents the new ellipse)
Best Answer
Consider an ellipse with center at the origin of course the foci will be at $(0,\pm{c})$ or $(\pm{c}, 0) $
As you have stated the eccentricity $e$=$\frac{c} {a}$ Note also that $c^2=a^2-b^2$, $c=\sqrt{a^2-b^2} $ where $a$ and $b$ are length of the semi major and semi minor axis and interchangeably depending on the nature of the ellipse
$e=\frac{c} {a}$ =$\frac{\sqrt{a^2-b^2}} {a}$=$\frac{\sqrt{a^2-b^2}} {\sqrt{a^2}}$
$e=\sqrt{\frac{a^2-b^2} {a^2} }$
Can you finish it from there?