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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.

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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{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$
answered Jun 19, 2013 by sreemathi.v
 

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