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The length of the latusrectum of $3 x^{2}-4 y+6 x-3=0$ is
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Verified Answer
The correct answer is:
$\frac{4}{3}$
Given equation of conic
$$
3 x^{2}-4 y+6 x-3=0
$$
$\begin{array}{lc}\Rightarrow & 3 x^{2}+6 x-3=4 y \\ \Rightarrow & 3\left(x^{2}+2 x-1\right)=4 y \\ \Rightarrow & 3\left(x^{2}+2 x+1-2\right)=4 y \\ \Rightarrow & 3(x+1)^{2}-6=4 y \\ \Rightarrow & 3(x+1)^{2}=4 y+6 \\ \text { Let } & (x+1)^{2}=\frac{4}{3}(y+3 / 2) \\ & X^{2}=\frac{4}{3} Y\end{array}$
$X^{2}=\frac{4}{3} Y$
where $X=x+1$ and $Y=y+3 / 2$
and $\quad 4 b=4 / 3 \Rightarrow b=1 / 3$
So, now the length of latusrectum is $4 / 3$.
$$
3 x^{2}-4 y+6 x-3=0
$$
$\begin{array}{lc}\Rightarrow & 3 x^{2}+6 x-3=4 y \\ \Rightarrow & 3\left(x^{2}+2 x-1\right)=4 y \\ \Rightarrow & 3\left(x^{2}+2 x+1-2\right)=4 y \\ \Rightarrow & 3(x+1)^{2}-6=4 y \\ \Rightarrow & 3(x+1)^{2}=4 y+6 \\ \text { Let } & (x+1)^{2}=\frac{4}{3}(y+3 / 2) \\ & X^{2}=\frac{4}{3} Y\end{array}$
$X^{2}=\frac{4}{3} Y$
where $X=x+1$ and $Y=y+3 / 2$
and $\quad 4 b=4 / 3 \Rightarrow b=1 / 3$
So, now the length of latusrectum is $4 / 3$.
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