logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

Find the equations of directrices, latus rectum and lengths of latus rectums of the following ellipses: $x^{2}+4y^{2}-8x-16y-68=0.$

Can you answer this question?
 
 

1 Answer

0 votes
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}$$=1$
  • http://clay6.com/mpaimg/11_2_Toolbox.png
  • Foci$(\pm ae,o),$ center $(0,0)$,vertices $(\pm a,0)$
  • End points of Latus Rectum $\; (ae,\pm \large\frac{b^2}{a})$ and $(-ae,\pm \large\frac{b^2}{a})$
  • Directrices $x=\pm \large\frac{a}{e}$.
  • The major axis is $y=0$ (x- axis) and the minor axes is $x=0$ (y- axis)
  • $\large\frac{x^2}{b^2}+\frac{y^2}{a^2}$$=1$
  • http://clay6.com/mpaimg/11_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:
$x^2-8x+4y^2-16y-68=0$
$x^2-8x+4y^2-16y=68$
Completing squares,
$(x^2-8x+16)+4(y^2-4y+4)=68+16+16$
$(x-4)^2+4(y-2)^2=100$
$\large\frac{(x-4)^2}{100}+\large\frac{(y-2)^2}{25}=1$
Step 2:
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}$$=1$
$a^2=100,b^2=25\Rightarrow a=10,b=5$
Step 3:
$e=\sqrt{1-\Large\frac{b^2}{a^2}}$
$\Rightarrow \sqrt{1-\large\frac{25}{100}}=\large\frac{\sqrt 3}{2}$
The major axis is the $X$-axis.
Step 4:
$XY$-axes :
Directrices $X=\pm\large\frac{a}{e}$
$X=\pm\large\frac{10}{\sqrt 3/2}$
$X=\pm \large\frac{20}{\sqrt 3}$
Latus rectum :
$X=\pm ae$
$X=\pm 10\times \large\frac{\sqrt 3}{2}$
$X=\pm 5\sqrt 3$
Length of $LR$ : $\large\frac{2b^2}{a}$
$\Rightarrow \large\frac{2\times 25}{10}$$=5$
Step 5:
$xy$-axes :
Directrices $x-4=\pm\large\frac{a}{e}$
$x-4=\pm\large\frac{10}{\sqrt 3/2}$
$x-4=\pm \large\frac{20}{\sqrt 3}$
$x=4\pm \large\frac{20}{\sqrt 3}$
Latus rectum :
$x-4=\pm 5\sqrt 3$
$x=4\pm 5\sqrt 3$
Length of $LR$ : $\large\frac{2b^2}{a}$
$\Rightarrow \large\frac{2\times 25}{10}$$=5$
answered Jun 17, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...