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

1 Answer

  • 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}$$=1$
  • Foci $(0,\pm ae)$,centre $(0,0)$,vertices $(0,\pm a)$.
  • Transverse axis $y$-axis ($x=0$)
  • Conjugate axis $x$-axis ($y=0$)
  • End points of latus rectum $(\pm\large\frac{b^2}{a},$$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 : $\large\frac{2b^2}{a}=\frac{2\times 4}{3}$
$\Rightarrow \large\frac{8}{3}$
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