Conic Sections — Maths STD 11 Science — Question

$\frac{x^2}{3}+\frac{y^2}{1}=1$

Comparing this equation with $\frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1$, we get

$\begin{aligned} & a^2=3 \text { and } b^2=1 \\ & a=\sqrt{3} \text { and } b=1\end{aligned}$

Since a > b,

X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2√3

Length of minor axis = 2b = 2(1) = 2

Lengths of the principal axes are 2√3 and 2.

(ii) We know that e $=\frac{\sqrt{a^2-b^2}}{a}$

$\begin{aligned} & =\frac{\sqrt{3-1}}{\sqrt{3}} \\ & =\frac{\sqrt{2}}{\sqrt{3}}\end{aligned}$

Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),

i.e., $S\left(\sqrt{ } 3\left(\frac{\sqrt{2}}{\sqrt{3}}\right), 0\right)$ and $S^{\prime}\left(-\sqrt{ } 3\left(\frac{\sqrt{2}}{\sqrt{3}}\right), 0\right)$

i.e., S(√2, 0) and S'(-√2, 0)

(iii) Equations of the directrices are $x= \pm \frac{a}{e}$,

$\begin{aligned} & = \pm \frac{\sqrt{3}}{\frac{\sqrt{2}}{\sqrt{3}}} \\ & = \pm \frac{3}{\sqrt{2}}\end{aligned}$

(iv) Length of latus rectum $=\frac{2 b^2}{a}=\frac{2(1)^2}{\sqrt{3}}=\frac{2}{\sqrt{3}}$

(v) Distance between foci = 2ae

$\begin{aligned} & =2(\sqrt{3})\left(\frac{\sqrt{2}}{\sqrt{3}}\right) \\ & =2 \sqrt{2}\end{aligned}$

(vi) Distance between directrices $=\frac{2 \mathrm{a}}{\mathrm{e}}$

$\begin{aligned} & =\frac{2 \sqrt{3}}{\frac{\sqrt{2}}{\sqrt{3}}} \\ & =\frac{2 \times 3}{\sqrt{2}} \\ & =3 \sqrt{2}\end{aligned}$