# Find the equation of the hyperbola if centre : $(1 , 4 );$one of the foci $(6 , 4 )$ and the corresponding directrix is $x=\large\frac{9}{4}$

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}$$=1 • http://clay6.com/mpaimg/toolbox_10(i).png • Foci (\pm ae,0),centre (0,0),vertices (\pm a,0). • Transverse axis x-axis (y=0) • Conjugate axis y-axis (x=0) • End points of latus rectum (ae,\pm\large\frac{b^2}{a}),($$-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:
$C(1,4)$ and $F(6,4)$ be on $y=4$ which is the transverse axis (parallel to $x$-axis)
Step 2:
$CF=ae$=5----(1)
$FZ=CF-CZ=ae-\large\frac{a}{e}$
$\Rightarrow 6-\large\frac{9}{4}=\frac{15}{4}$----(2)
$\Rightarrow \large\frac{a}{e}=$$5-\large\frac{15}{4}=\frac{5}{4}-----(3) Step 3: Eq(1) & Eq(2) gives a^2=\large\frac{25}{4}$$\Rightarrow a=\large\frac{5}{2}$
$e=\large\frac{5}{a}=\large\frac{5}{5/2}=$$2 b=a\sqrt{e^2-1}=\large\frac{5}{2}$$\sqrt{4-1}$$=\large\frac{5\sqrt 3}{2} Step 4: Since the transverse axis is parallel to the x-axis the equation is of the form \large\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}$$=1$
Where $(h,k)$ is the centre.