# Find the equation of the hyperbola if foci: $(\pm 3 , 5 ); e=3$

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}$$=1 • http://clay6.com/mpaimg/2_toolbox_10(i).png • Foci (\pm ae,0),centre (0,0),vertices (\pm a,0). • Transverse axis x-axis (y=0) • Conjugate axis y-axis (x=0) • End points of latus rectum (ae,\pm\large\frac{b^2}{a}),($$-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:
Distance between the foci =$2ae=6$
$\Rightarrow ae=3$
Now $e=3$
Therefore $3a=3$
$a=1$
$b=a\sqrt{e^2-1}$
$\Rightarrow \sqrt{9-1}=2\sqrt 2$
Step 2:
The transverse axis is $y=5$(parallel to $x$-axis)
Since it contains the foci.
The centre is the mid point of $FF'$(i.e) $C(0,5)$
Step 3:
The equation is of the form
$\large\frac{(x-h)^2}{a^2}-\large\frac{(y-k)^2}{b^2}$$=1 Where (h,k) is (0,5) \large\frac{x^2}{1}-\large\frac{(y-5)^2}{8}$$=1$
edited Jun 19, 2013