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Find the equations of directrices, latus rectum and lengths of latus rectums of the following ellipses: $3x^{2}+2y^{2}-30x-4y+23=0$

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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 $e^2=1-\large\frac{b^2}{a^2}$
  • 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}$$=1$
  • http://clay6.com/mpaimg/12_2_Toolbox.png
  • Foci$(\pm ae,o),$ center $(0,0)$,vertices $(\pm a,0)$
  • End points of Latus Rectum $\; (ae,\pm \large\frac{b^2}{a})$ and $(-ae,\pm \large\frac{b^2}{a})$
  • Directrices $x=\pm \large\frac{a}{e}$.
  • The major axis is $y=0$ (x- axis) and the minor axes is $x=0$ (y- axis)
  • $\large\frac{x^2}{b^2}+\frac{y^2}{a^2}$$=1$
  • http://clay6.com/mpaimg/12_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}$$=1$
  • In this case major axis y=k and minor axis $x=h$.
  • $\large\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}$$=1$
  • In this case major axis $x=h$ and minor axis $y=k$.
Step 1:
$3x^2-30x+2y^2-4y+23=0$
$3x^2-30x+2y^2-4y=-23$
Completing the squares
$3(x^2-10x+25)+2(y^2-2y+1)=-23+75+2$
$3(x-5)^2+2(y-1)^2=54$
$\large\frac{(x-5)^2}{18}+\frac{(y-1)^2}{27}$$=1$
$a^2=27,b^2=18$
$\Rightarrow a=3\sqrt 3,b=3\sqrt 2$
Step 2:
$e=\sqrt{1-\large\frac{b^2}{a^2}}$
$\Rightarrow \sqrt{1-\large\frac{18}{27}}=\large\frac{1}{\sqrt 3}$
Step 3:
Shifting the origin $E(5,1)$ by translation of axes $X=x-5$(or $x=X+5)$,$Y=y-1$ (or $y=Y+1)$
$\large\frac{X^2}{18}+\frac{Y^2}{27}$$=1$
The $Y$-axis is the major axis.
Step 4:
$XY$-axes :
Directrices $Y=\pm\large\frac{a}{e}$
$Y=\pm \large\frac{3\sqrt 3}{1/\sqrt 3}$
$Y=\pm 9$
Latus rectum : $Y=\pm ae$
$\Rightarrow Y=\pm 3\sqrt 3\large\frac{1}{\sqrt 3}$
$\Rightarrow Y=\pm 3$
Step 5:
$xy$-axes :
$y-1=\pm 9$
$y=10,y=-8$
Latus rectum : $y-1=\pm 3$
$\Rightarrow y=4,y=-2$
Step 6:
Length of LR : $\large\frac{2b^2}{a}$
$\Rightarrow \large\frac{2\times 18}{3\sqrt 3}$
$\Rightarrow \large\frac{12}{\sqrt 3}=$$4\sqrt 3$
answered Jun 18, 2013 by sreemathi.v
 

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