Gujarat BoardEnglish MediumSTD 11 ScienceMATHSCONIC SECTIONS2 Marks
Question
Find the equation of hyperbola having Foci (0, $\pm$13) and the conjugate axis is of length 24.
✓
Answer
Here foci are $(0, \pm 13)$ which lie on $y$-axis.
So the equation of hyperbola in standard form is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$\therefore(13)^2=a^2+(12)^2$
$\Rightarrow a^2=169-144=25$
Thus required equation of hyperbola is
$\frac{y^2}{25}-\frac{x^2}{(12)^2}=1 \Rightarrow \frac{y^2}{25}-\frac{x^2}{144}=1$
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