Conic Sections — Maths STD 11 Science — Question

$\frac{x^2}{\frac{1}{3}}+\frac{y^2}{\frac{1}{4}}=1$

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

$\begin{aligned} & a^2=\frac{1}{3} \text { and } b^2=\frac{1}{4} \\ & a=\frac{1}{\sqrt{3}} \text { and } b=\frac{1}{2}\end{aligned}$

Since a > b,

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

(i) Length of major axis $=2 a=2\left(\frac{1}{\sqrt{3}}\right)=\frac{2}{\sqrt{3}}$

Length of minor axis $=2 b=2\left(\frac{1}{2}\right)=1$

Lengths of the principal axes are $\frac{2}{\sqrt{3}}$ and 1 .

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

$e=\frac{\sqrt{\frac{1}{3}-\frac{1}{4}}}{\frac{1}{\sqrt{3}}}=\frac{\sqrt{\frac{1}{12}}}{\frac{1}{\sqrt{3}}}=\sqrt{\frac{3}{12}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$

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

$\begin{aligned} & \text { i.e. } S\left(\frac{1}{\sqrt{3}}\left(\frac{1}{2}\right), 0\right) \text { and } S^{\prime}\left(-\frac{1}{\sqrt{3}}\left(\frac{1}{2}\right), 0\right) \\ & \text { i.e., } S\left(\frac{1}{2 \sqrt{3}}, 0\right) \text { and } S^{\prime}\left(-\frac{1}{2 \sqrt{3}}, 0\right)\end{aligned}$

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

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

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

$\begin{aligned} & =\frac{2\left(\frac{1}{2}\right)^2}{\frac{1}{\sqrt{3}}} \\ & =\frac{\sqrt{3}}{2}\end{aligned}$

$\begin{aligned} & =2\left(\frac{1}{\sqrt{3}}\right)\left(\frac{1}{2}\right) \\ & =\frac{1}{\sqrt{3}}\end{aligned}$

(v) Distance between foci = 2ae

$\begin{aligned} & =2\left(\frac{1}{\sqrt{3}}\right)\left(\frac{1}{2}\right) \\ & =\frac{1}{\sqrt{3}}\end{aligned}$

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

$\begin{aligned} & =\frac{2\left(\frac{1}{\sqrt{3}}\right)}{\frac{1}{2}} \\ & =\frac{4}{\sqrt{3}}\end{aligned}$