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