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# Find the equation and the length of the transverse and conjugate axis of the following hyperbola:$16x^{2}-9y^{2}+96x+36y-36=0$

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{x^2}{a^2}-\frac{y^2}{b^2}$$=1 • http://clay6.com/mpaimg/4_toolbox_10(i).png • Foci (\pm ae,0),centre (0,0),vertices (\pm a,0). • Transverse axis x-axis (y=0) • Conjugate axis y-axis (x=0) • End points of latus rectum (ae,\pm\large\frac{b^2}{a}),($$-ae,\pm\large\frac{b^2}{a})$
• 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:
$16x^2-9y^2+96x+36y-36=0$
$16x^2+96x-9y^2+36y=36$
Completing squares,
$16(x^2+6x+9)-9(y^2-4y+4)=36+144-36$
$16(x+3)^2-9(y-2)^2=144$
Step 2:
The above equation is divided by $144$
$\large\frac{(x+3)^2}{9}-\frac{(y-2)^2}{16}$$=1 a^2=9,b^2=16 \Rightarrow a=3,b=4 Step 3: Shifting the origin to (-3,-2) by translation of axes X=x+3,Y=y-2 \Rightarrow x=X-3,y=Y+2 The equation reduces to \large\frac{X^2}{9}-\frac{Y^2}{16}$$=1$
Transverse axes : $X$-axes (i.e) $Y=0\Rightarrow y-2=0$
Conjugate axes : $Y$-axes (i.e) $X=0\Rightarrow x+3=0$
Length of transverse axis =$2a=2\times 3=6$
Length of conjugate axis =$2b=2\times 4=8$
edited Jun 19, 2013