Question

Find the co-ordinates of the foci, the vertices, the length of major axis, latus rectum and the eccentricity of the conic represented by the equation $3x^2+5y^2=15$

Answer 1

The equation of the given conic is $3x^2+5y^2=15$

It can be written as $\frac{x^2}{5}+\frac{y^2}{3}=1$ ........$\left(1\right)$

which is comparable with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

So $\left(1\right)$ represents an ellipse

Here $a^2=5,b^2=3\Rightarrow a=\sqrt{5},b=\sqrt{3}$

We know that $c=\sqrt{a^2-b^2}=\sqrt{5-3}=\sqrt{2}$

$\therefore$ the coordinates of foci are $(-c,0)$ and $(c,0)$

i.e.,$(-\sqrt{2},0)$ and $(\sqrt{2},0)$

The co-ordinate of vertices are $(-a,0)$ and $(a,0)$

i.e.,$(-\sqrt{5},0)$ and $(\sqrt{5},0)$

Length of major axis$=2a=2\times \sqrt{5}=2\sqrt{5}$

Length of minor axis$=2b=2\times \sqrt{3}=2\sqrt{3}$

Length of latus rectum$=\frac{2b^2}{a}=\frac{2\times {\left(\sqrt{3}\right)}^2}{\sqrt{5}}=\frac{2\times 3}{\sqrt{5}}=\frac{6}{\sqrt{5}}$

Eccentricity$=\frac{c}{a}=\frac{\sqrt{2}}{\sqrt{5}}=\sqrt{\frac{2}{5}}$