The ellipses have their centers at the origin and their major axes on the $x$-axis. Find the equation:
- with distance between directrices $27$, and between foci $3$;
- with a focus at $(-\sqrt{13},0)$ and a vertex at $(0,2)$.
[Math] Find equations of the ellipses given conditions on the directrices, foci, and vertices
analytic geometryconic sections
Related Solutions
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)
If $a$ is the semi-major axis of one such ellipse, and $F$ its other focus, then $$ |R_1|+|R_1-F|=2a=|R_2|+|R_2-F|. $$ Subtracting one from the other, $F$ must satisfy the following equation: $$ |F-R_1|-|F-R_2|=|R_2|-|R_1|. $$ The set of all points $F$ that satisfy this equation is a hyperbola with focuses $R_1$ and $R_2$, and the semi-major axis $-\frac12\big||R_1|-|R_2|\big|$, and it is easy to find coordinates of points on the hyperbola using standard formulas (see the mathworld link, e.g.).
Best Answer
The equation of an ellipse with the center at the origin and the major axes on the x-axis is $$\frac {x^2}{a^2}+\frac {y^2}{b^2}=1$$
where $2a,2b$ are the major & minor axes respectively.
We know the coordinate of the foci are $(\pm ae,0)$ and the equation of directrices are $x=\pm\frac ae$ where $e$ is the Eccentricity $e=\frac{\sqrt{a^2-b^2}}b$
So the distance between the foci is $2ae\implies 2ae=3$ and
the distance between the directrices is $\frac{2a}e\implies \frac{2a}e=27$
Solve for $a,e$
$2.$ As the coordinate of the foci are $(\pm ae,0), ae=\sqrt{13}$ as $ae>0$
The vertices are $(\pm a,0)$ and $(0,\pm b)\implies b=2$ as $b>0$
As $b^2=a^2(1-e^2)\iff a^2=b^2+(ae)^2=2^2+13=17$