Question

The length of the latus-rectum of the conic $3x^2+4y^2-6x+8y-5=0$ is unit.

Answer 1

$3x^2+4y^2-6x+8y-5=0$
$\Rightarrow 3(x^2-2x)+4(y^2+2y)=5$
$\Rightarrow 3(x^2-2x+1)+4(y^2+2y+1)=5+3+4$
$\Rightarrow 3(x-1{)}^2+4(y+1{)}^2=12$
$\Rightarrow \frac{(x-1{)}^2}{4}+\frac{(y+1{)}^2}{3}=1$
So, $a=2$ and $b=\sqrt{3}(a>b).$

$\therefore$ Latus rectum $=\frac{2b^2}{a}$
$=\frac{2\times 3}{2}$
$=3$ unit