Q91 / 180 · KCET 2011 · Mathematics

The length of the latusrectum of $3 x^{2}-4 y+6 x-3=0$ is

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

Solution

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 $4 / 3$.

Hence, the correct answer is option: 'B'.

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