Question

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

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

Given equation of conic
$3x^2-4y+6x-3=0$
$\Rightarrow 3x^2+6x-3=4y$
$\Rightarrow 3\left(x^2+2x-1\right)=4y$
$\Rightarrow 3\left(x^2+2x+1-2\right)=4y$
$\Rightarrow 3(x+1{)}^2-6=4y$
$\Rightarrow 3(x+1{)}^2=4y+6$
$\Rightarrow (x+1{)}^2=\frac{4}{3}(y+3/2)$...(i)
Let $X^2=\frac{4}{3}Y$...(ii)
where $X=x+1$ and $Y=y+3/2$
and $4b=4/3\Rightarrow b=1/3$
So, now the length of latusrectum is $4/3.$