# Find the equation of the hyperbola if centre : $(1 , -2 )$ length of the transverse axis is $8; e=\large\frac{5}{4}$ and the transverse axis is parallel to x- axis.

## 1 Answer

Toolbox:
• 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:
$C(1,-2),2a=8,e=\large\frac{5}{4}$
Transverse axis is parallel to $x$-axis.
Therefore transverse axis is $y=-2$
$a=4,b^2=a^2(e^2-1)$
$\Rightarrow 16\big(\large\frac{25}{16}$$-1)=9 Step 2: Equation is of the form \large\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}$$=1$
$\Rightarrow \large\frac{(x-1)^2}{16}-\frac{(y-(-2))^2}{9}$$=1 The equation is \large\frac{(x-1)^2}{16}-\frac{(y+2)^2}{9}$$=1$
answered Jun 19, 2013

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