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# Find the equation of the hyperbola if foci: $(6 , -1 )$ and $(-4 , -1)$ and passing through the point $(4 , -1 )$

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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}$$=1 • http://clay6.com/mpaimg/1_toolbox_10(i).png • Foci (\pm ae,0),centre (0,0),vertices (\pm a,0). • Transverse axis x-axis (y=0) • Conjugate axis y-axis (x=0) • End points of latus rectum (ae,\pm\large\frac{b^2}{a}),($$-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:
The centre is the midpoint of $FF'$.
(i.e) $C(\large\frac{6-4}{2},$$-1)=(1,-1) The transverse axis contains the foci.Its equation is y=-1 (parallel to x-axis). The distance between the foci FF'=2ae=10 \Rightarrow ae=5---(1) Step 2: The point P(4,-1) lies on the curve. It lies on the transverse axis y=-1 It is a vertex of the hyperbola. Therefore CA=a=3(length of semi transverse axis). Step 3: Substituting a=3 in (1),e=\large\frac{5}{3} Therefore b=\sqrt{\large\frac{25}{9}-\normalsize 1} \Rightarrow b=3\times \large\frac{4}{3}$$=4$
Step 4:
The equation is of the form
$\large\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}$$=1 Where (h,k) is C(1,-1) Therefore the equation is \large\frac{(x-1)^2}{9}-\frac{(y+1)^2}{16}$$=1$
answered Jun 19, 2013
edited Jun 19, 2013