# Find the eccentricity, centre, foci, vertices of the following ellipses and draw the diagram:$x^{2}+4y^{2}-8x-16y-68=0$

Toolbox:
• Shifting of the origin from $(0,0)$ to $(h,k)$ then if the new axes form $X-Y$ coordinate system then $x=X+a$,$y=Y+k$
• The new coordinates of a point $P(x,y)$ become $X=x-a,Y=y-k$
• Standard forms of equation of the ellipse with major axes 2a, minor axis 2b $(a >b)$ eccentricity e and $b^2=a^2(1-e^2)$ or $e^2=1-\large\frac{b^2}{a^2}$
• $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1 • http://clay6.com/mpaimg/14_2_Toolbox.png • Foci(\pm ae,o), center (0,0),vertices (\pm a,0) • End points of Latus Rectum \; (ae,\pm \large\frac{b^2}{a}) and (-ae,\pm \large\frac{b^2}{a}) • Directrices x=\pm \large\frac{a}{e}. • The major axis is y=0 (x- axis) and the minor axes is x=0 (y- axis) • \large\frac{x^2}{b^2}+\frac{y^2}{a^2}$$=1$
• http://clay6.com/mpaimg/14_2_Toolbox1.png
• Foci$(0,\pm ae),$ center $(0,0)$,vertices $(0,\pm a)$
• End points of Latus Rectum $(\pm \large\frac{b^2}{a}$$,ae) and (\pm \large\frac{b^2}{a}$$,-ae)$
• Directrices $y=\pm \large\frac{a}{e}$.
• The major axis is $x=0$ (y- axis) and the minor axes is $y=0$ (x- axis)
• General form of standard ellipses with centre $C$ with major axis $2a$,minor axis $2b$ and axes parallel to the coordinate axes.
• $\large\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}$$=1 • In this case major axis y=k and minor axis x=h. • In this case major axis x=h and minor axis y=k. Step 1: x^2+4y^2-8x-16y-68=0 x^2-8x+4y^2-16y=68 (x^2-8x+16)+4(y^2-4y+4)=68+16+16 (x-4)^2+4(y-2)^2=100 The above equation is divided by 100 Step 2: \large\frac{(x-4)^2}{100}+\frac{(y-2)^2}{25}$$=1$
Shifting the origin to $(4,2)$ by translation of axes.
$X=x-4$ (or $x=X+4)$
$Y=y-2$ (or $Y=y+2$)
$\large\frac{X^2}{100}+\frac{Y^2}{25}$$=1 a^2=100,b^2=25 \Rightarrow a=10,b=5. Step 3: The major axes is the X-axes.(i.e) Y=0 or y-2=0 Step 4: XY-axes : Eccentricity e=1-\sqrt{\large\frac{b^2}{a^2}} \Rightarrow \sqrt{1-\large\frac{25}{100}}=\sqrt{\frac{75}{100}}=\large\frac{\sqrt3}{2} Centre : (0,0) Foci : (\pm ae,0) ae=10\times \large\frac{\sqrt 3}{2}=$$5\sqrt 3$
$\Rightarrow (\pm 5\sqrt 3,0)$
Vertices : $(\pm a,0)$
$\Rightarrow (\pm 10,0)$
Step 5:
$xy$-axes :
Centre : $(4,2)$
Foci : $(\pm 5\sqrt 3+4,2)$
Vertices : $(\pm 10+4,2)$
$\Rightarrow (14,2),(-6,2)$
edited Jun 21, 2013