Question

Find the centre, the lengths of the axes, eccentricity, foci of the ellipse:
$\left(iv\right)3x^2+4y^2-12x-8y+4=0$

Answer 1

Given: $3x^2+4y^2-12x-8y+4=0$

$\Rightarrow 3(x^2-4x+4)+4(y^2-2y+1)-12=0$

$\Rightarrow 3(x-2{)}^2+4(y-1{)}^2=12$

$\Rightarrow \frac{(x-2{)}^2}{4}+\frac{(y-1{)}^2}{3}=1$

Comparing with $\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1$

Where, centre $=(h,k)=(2,1)$

$a^2=4\Rightarrow a=2$

$b^2=3\Rightarrow b=\sqrt{3}$

Length of major axis $=2a=4$

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

Eccentricity, $e=\sqrt{1-\frac{b^2}{a^2}}$

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

Foci $=(h\pm ae,k)=(2\pm 1,1)$

Hence,

Centre $=(2,1)$

Length of Major axis $=4$

Length of Minor axis $=2\sqrt{3}$

Eccentricity $=\frac{1}{2}$

Foci $=(2\pm 1,1)$