# Find the equation of the hyperbola if centre:$(2 , 5 )$; the distance between the direction is $15$ the distance between the foci is $20$ and the transverse axis is parallel to y-axis.

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}$$=1 • http://clay6.com/mpaimg/1_toolbox10.jpg • Foci (0,\pm ae),centre (0,0),vertices (0,\pm a). • Transverse axis y-axis (x=0) • Conjugate axis x-axis (y=0) • End points of latus rectum (\pm\large\frac{b^2}{a},$$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}$$=1 • Transverse axis x-h=0,conjugate axis y-k=0. Step 1: C(2,5),Distance between directrices 15,distance between foci 20,transverse axes parallel to y-axis. Distance between directrices =\large\frac{2a}{e}$$=15$
$\Rightarrow \large\frac{a}{e}=\frac{15}{2}$------(1)
Distance between foci $=2ae=20$
$\Rightarrow ae=10$-------(2)
Eq(1)&Eq(2) gives $a^2=75$
(i.e)$a=5\sqrt 3$
Therefore $e=\large\frac{10}{a}=\frac{10}{5\sqrt 3}=\frac{2}{\sqrt 3}$
$b=a\sqrt{e^2-1}=5\sqrt 3\sqrt{\large\frac{2}{3}-\normalsize 1}=5$
Step 2:
Equation is of the form $\large\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}$$=1 Where (h,k) is the centre. Therefore the equation is \large\frac{(y-5)^2}{75}-\frac{(x-2)^2}{25}$$=1$