Maharashtra BoardEnglish MediumSTD 11 ScienceMathsConic Sections4 Marks
Question
$\frac{x^2}{144}-\frac{y^2}{25}=1$
✓
Answer
Given equation of the hyperbola $\frac{x^2}{144}-\frac{y^2}{25}=1$Comparing this equation with $\frac{x^2}{a^2}-\frac{y^2}{h^2}=1$
$a^2=144 \text { and } b^2=25$
$\because a=12 \text { and } b=5$
(i) Length of transverse axis = 2a = 2(12) = 24 Length of conjugate axis = 2b = 2(5) = 10 lengths of the principal axes are 24 and 10.
$25=144\left( e ^2-1\right)$
$\frac{25}{144}= e ^2-1$
$e ^2=1+\frac{25}{144}$
$e ^2=\frac{169}{144}$
$e =\frac{13}{12} \ldots \ldots .[\because e >1]$
Co-ordinates of foci are S(ae, 0) and S'(-ae, 0)
i.e., $S\left(12\left(\frac{13}{12}\right), 0\right)$ and $S^{\prime}\left(-12\left(\frac{13}{12}\right), 0\right)$
i.e., $S(13,0)$ and $S^{\prime}(-13,0)$
(iii) Equations of the directrices are $x= \pm \frac{a}{e}$
$\text { i.e., } x= \pm \frac{12}{\left(\frac{13}{12}\right)}$
$\text { i.e., } x= \pm \frac{144}{13}$
(iv) Length of latus rectum $=\frac{2 b^2}{a}=\frac{2(25)}{12}=\frac{25}{6}$
(v) Distance between foci $=2 ae =2(12)\left(\frac{13}{12}\right)=26$
(vi) Distance between directrices $=\frac{2 a }{ e }=\frac{2(12)}{\left(\frac{13}{12}\right)}=\frac{288}{13}$
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