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10-2. Parabola, Ellipse, Hyperbola
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प्रतिबंधों को संतुष्ट करते हुए अतिपरवलय का समीकरण ज्ञात कीजिए
शीर्ष $(\pm 2,0),$ नाभियाँ $(±3,0)$
A
$\frac{x^{2}}{4}-\frac{y^{2}}{5}=1$
B
$\frac{x^{2}}{4}-\frac{y^{2}}{5}=1$
C
$\frac{x^{2}}{4}-\frac{y^{2}}{5}=1$
D
$\frac{x^{2}}{4}-\frac{y^{2}}{5}=1$
Solution
Vertices $(\pm 2,\,0),$ foci $(±3,\,0)$
Here, the vertices are on the $x-$ axis.
Therefore, the equation of the hyperbola is of the form $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
since the vertices are $(\pm 2,\,0)$, $a =2$
since the foci are $(\pm 3,\,0)$, $c=3$
We know that $a^{2}+b^{2}=c^{2}$
$\therefore 2^{2}+b^{2}=3^{2}$
$b^{2}=9-4=5$
Thus, the equation of the hyperbola is $\frac{x^{2}}{4}-\frac{y^{2}}{5}=1$
Standard 11
Mathematics
