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

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A)
Toolbox:
• 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}$$=1 • Transverse axis x-h=0,conjugate axis y-k=0. Step 1: 9x^2-4y^2-36x+32y+8=0 9x^2-4y^2-36x+32y=-8 Completing squares, 9(x^2-4x+4)-4(y^2-8y+16)=-8+36-64 9(x-2)^2-4(y-4)^2=-36 Step 2: 4(y-4)^2-9(x-2)^2=36 Completing squares, \large\frac{(y-4)^2}{9}-\frac{(x-2)^2}{4}$$=1$
$a^2=9,b^2=4$
$\Rightarrow a=3,b=2$
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 :
$Y=\pm\large\frac{a}{e}$
$Y=\pm\large\frac{3}{\sqrt{13}/3}$
$Y=\pm\large\frac{9}{\sqrt{13}}$
$y=4\pm\large\frac{9}{\sqrt{13}}$
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 : $\large\frac{2b^2}{a}=\frac{2\times 4}{3}$
$\Rightarrow \large\frac{8}{3}$