Question

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

• A. 3
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{2}{\sqrt{3}}$
• D. none of these

Answer 1

The correct option is A 3

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