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Find the equations of directrices, latus rectum and lengths of latus rectums of the following ellipses: $25x^{2}+169y^{2}=4225$.

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Toolbox:
  • 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}$$=1$
  • http://clay6.com/mpaimg/9_2_Toolbox.png
  • Foci$(\pm ae,o),$ center $(0,0)$,vertices $(\pm a,0)$
  • End points of Latus Rectum $\; (ae,\pm \large\frac{b^2}{a})$ and $(-ae,\pm \large\frac{b^2}{a})$
  • Directrices $x=\pm \large\frac{a}{e}$.
  • The major axis is $y=0$ (x- axis) and the minor axes is $x=0$ (y- axis)
  • $\large\frac{x^2}{b^2}+\frac{y^2}{a^2}$$=1$
  • http://clay6.com/mpaimg/9_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)
Step 1:
$25x^2+169y^2=4225$
Dividing the above equation by $4225$ we get
$\large\frac{x^2}{169}+\frac{y^2}{25}=$$1$
$a^2=169,b^2=25\Rightarrow a=13,b=5$
The major axis is the $x$-axis, $y=0$.
Step 2:
Eccentricity $e=\sqrt{1=\large\frac{b^2}{a^2}}$
$\Rightarrow \sqrt{1-\large\frac{25}{169}}=\large\frac{12}{13}$
Step 3:
Directrices $x=\pm\large\frac{a}{e}$
$x=\pm \large\frac{13}{\Large\frac{12}{13}}\Rightarrow \pm\large\frac{169}{12}$
Step 4:
Latus rectum $x=\pm ae$
$x=\pm 13.\large\frac{12}{13}$
$x=\pm 12$
Step 5:
Length of $LR=\large\frac{2b^2}{a}$
$\Rightarrow \large\frac{2\times 25}{13}=\frac{50}{13}$
answered Jun 17, 2013 by sreemathi.v
 

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