Conic Sections — Maths STD 11 Science — Question

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

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

$a^2=4$ and $b^2=3$

a = 2 and b = √3

Since a > b,

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

(i) Length of major axis = 2a = 2(2) = 4

Length of minor axis = 2b = 2√3

Lengths of the principal axes are 4 and 2√3.

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

$\begin{aligned} & =\frac{\sqrt{4-3}}{2} \\ & =\frac{1}{2}\end{aligned}$

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

i.e., $S\left(2\left(\frac{1}{2}\right), 0\right)$ and $S^{\prime}\left(-2\left(\frac{1}{2}\right), 0\right)$

i.e., S(1, 0) and S'(-1, 0)

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

$\begin{aligned} & = \pm \frac{2}{\frac{1}{2}} \\ & = \pm 4\end{aligned}$

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

(v) Distance between foci $=2 a e=2(2)\left(\frac{1}{2}\right)=2$

$\begin{aligned} & =\frac{2(2)}{\frac{1}{2}} \\ & =8\end{aligned}$