Ask Questions, Get Answers

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

Find the equations of directrices ,latus rectums and length of latus rectum of the following hyperbolas:$9x^{2}-4y^{2}-36x+32y+8=0$

Can you answer this question?

1 Answer

0 votes
  • Standard forms of equation of the hyperbola with transverse axis $2a$,conjugate axis $2t$ with the negative sign associated with $b$ and $e=\sqrt{1+\large\frac{b^2}{a^2}},b=a\sqrt{e^2-1}$
  • $\large\frac{y^2}{a^2}-\frac{x^2}{b^2}$$=1$
  • http://clay6.com/mpaimg/4_toolbox10.jpg
  • Foci $(0,\pm ae)$,centre $(0,0)$,vertices $(0,\pm a)$.
  • Transverse axis $y$-axis ($x=0$)
  • Conjugate axis $x$-axis ($y=0$)
  • End points of latus rectum $(\pm\large\frac{b^2}{a},$$ae),(\pm\large\frac{b^2}{a},$$-ae)$
  • Length of LR :$\large\frac{2b^2}{a}$
  • Directrices $y=\pm\large\frac{a}{e}$
  • General form of standard hyperbola with centre $C(h,k)$,transverse axis $2a$,conjugate axis $2b$,$(b^2-ve)$ and with axes parallel to the coordinate axes.
  • (i) $\large\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}$$=1$
  • Transverse axis $y-k=0$,conjugate axis $x-h=0$.
  • (ii) $\large\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}$$=1$
  • Transverse axis $x-h=0$,conjugate axis $y-k=0$.
Step 1:
Completing squares,
Step 2:
Completing squares,
$\Rightarrow a=3,b=2$
Step 3:
eccentricity $e=\sqrt{1+\large\frac{b^2}{a^2}}$
$\Rightarrow \sqrt{1+\large\frac{4}{9}}=\large\frac{\sqrt{13}}{3}$
Step 4:
Shifting the origin to $(2,y)$ by translation of ads $Y=y-4,X=x-2$
(i.e) $y=Y+4,x=X+2$
The equation reduces to $\large\frac{Y^2}{9}-\frac{X^2}{4}$$=1$
The transverse axis is the $Y$-axis .
(i.e) $X=0$ or $x-2=0$
Step 5:
Directrices :
Step 6:
Latus rectum : $Y=\pm ae$
$Y=\pm 3\times \large\frac{\sqrt{13}}{3}$
$Y=\pm \sqrt{13}$
$y=4\pm \sqrt{13}$
Step 7:
Length of LR : $\large\frac{2b^2}{a}=\frac{2\times 4}{3}$
$\Rightarrow \large\frac{8}{3}$
answered Jun 19, 2013 by sreemathi.v

Related questions

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