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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Find the equations of the hyperbola satisfying the given conditions : Foci $(0, \pm \sqrt{ 10} )$ , passing through $(2,3 )$

$\begin {array} {1 1} (A)\;x^2-y^2=5 & \quad (B)\;y^2-x^2=5 \\ (C)\;x^2+y^2=5 & \quad (D)\; \text{None of the above} \end {array}$

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Toolbox:
  • Standard equation of a hyperbola where transverse axis is along x - axis is $ \large\frac{x^2}{a^2}$$-\large\frac{y^2}{b^2} $$=1$
  • Standard equation of a hyperbola whose transverse axis is along y - axis is $ \large\frac{y^2}{a^2} $$ - \large\frac{x^2}{b^2}$$=1$
  • $c^2=a^2+b^2$, where $c$ is the foci
  • Eccentricity $e = \large\frac{c}{a}$
  • Length of the latus rectum $ = \large\frac{2b^2}{a}$
Step 1 :
It is given that the coordinates of foci are $( 0, \pm \sqrt{10})$
The hyperbola passes through the point (2, 3)
It is clear that the foci lie on the y - axis.
Hence the equation of the hyperbola should be of the form $ \large\frac{y^2}{a^2} - \large\frac{x^2}{b^2}=1$
Since the foci are $(0, \pm \sqrt{10})$
$c=\sqrt{10}$
$c^2=a^2+b^2$
Step 2 :
$c^2=a^2+b^2$
(i.e) $10 = a^2+b^2$ or $ b^2=10-a^2$
Step 3 :
Since the hyperbola passes through the point (2, 3)
$ \large\frac{9}{a^2} - \large\frac{4}{b^2}$$ =1$ (i.e) $9b^2-4a^2=a^2b^2$
Substituting for $b^2$ we get,
$ \Rightarrow 9(10-a^2)-4a^2=a^2(10-a^2) \quad $(i.e) $90-9a^2-4a^2=10a^2-a^4$
On rearranging we get,
$a^4-23a^2+90=0$
On factorizing we get,
$a^4-18a^2-5a^2+90=0$
$a^2(a^2-18)-5(a^2-18)=0$
(i.e) $(a^2-5)(a^2-18)=0$
$ \therefore a = 5$ or 18
Step 4 :
But in a hyperbola $c > a$
(i.e) $c^2 > a^2$
(i.e) $a^2=18$ is not admissable
$ \therefore b^2=10-5=5$
Hence the equation of the hyperbola is
$ \large\frac{y^2}{5} - \large\frac{x^2}{5}=1$
or $y^2-x^2=5$
answered Jul 21, 2014 by thanvigandhi_1
 

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