Question

find the equation of hyperbola whose traverse and conjugate axes are the x and y axes respectivlygiven that the length of conjugate axis 5 and distance between the foci is 13.

Answer 1

$$\begin{gathered} \mathrm{Given}:-\mathrm{Length}\mathrm{of}\mathrm{conjugate}\mathrm{axes}=5\mathrm{unit}=2\mathrm{b}. \\ \Rightarrow \mathrm{b}=\frac{5}{2}\mathrm{unit}. \\ \mathrm{Distance}\mathrm{between}\mathrm{the}\mathrm{foci}=13\mathrm{unit}=2\mathrm{c} \\ \Rightarrow \mathrm{c}=\frac{13}{2}\mathrm{unit}. \\ \mathrm{We}\mathrm{know}, \\ \mathrm{a}=\sqrt{{\mathrm{c}}^2-{\mathrm{b}}^2}=\sqrt{{\left(\frac{13}{2}\right)}^2-{\left(\frac{5}{2}\right)}^2} \\ =\sqrt{\frac{169-25}{4}}=\sqrt{\frac{144}{4}} \\ =\sqrt{36}=6\mathrm{unit}. \\ \mathrm{As}\mathrm{transverse}\mathrm{axis}\mathrm{is}\mathrm{X},\mathrm{therefore}\mathrm{hyperbola}\mathrm{is}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \\ \mathrm{So},\mathrm{equaion}\mathrm{of}\mathrm{hyperbola}\mathrm{is} \\ \frac{{\mathrm{x}}^2}{{\left(6\right)}^2}-\frac{{\mathrm{y}}^2}{{\left({\displaystyle \frac{5}{2}}\right)}^2}=1 \\ \frac{{\mathrm{x}}^2}{36}-\frac{4{\mathrm{y}}^2}{25}=1. \end{gathered}$$