Question

Find the centre, the lengths of the axes, eccentricity, foci of the ellipse:
$\left(v\right)4x^2+16y^2-24x-32y-12=0$

Answer 1

Given: $4x^2+16y^2-24x-32y-12=0$

$\Rightarrow 4(x^2-6x+9)+16(y^2-2y+1)-64=0$

$\Rightarrow 4(x-3{)}^2+16(y-1{)}^2=64$

$\Rightarrow \frac{(x-3{)}^2}{16}+\frac{(y-1{)}^2}{4}=1$

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

Where, centre $=(h,k)=(3,1)$

$a^2=16\Rightarrow a=4$

$b^2=4\Rightarrow b=2$

Length of major axis $=2a=8$

Length of minor axis $=2b=4$

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

$\Rightarrow e=\sqrt{1-\frac{4}{16}}=\frac{\sqrt{3}}{2}$

Foci $=(h\pm ae,k)=(3\pm 2\sqrt{3},1)$

Hence,
Centre $=(3,1)$

Length of Major axis $=8$

Length of Minor axis $=4$

Eccentricity $=\frac{\sqrt{3}}{2}$

Foci $=(3\pm 2\sqrt{3},1)$