Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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

Can you answer this question?

1 Answer

0 votes
  • 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
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 by sreemathi.v
edited Jun 19, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App