The distance between the vertex and the focus of the parabola $x^2-2 x+3 y-2=0$ is

The distance between the vertex and the focus of the parabola $x^2-2 x+3 y-2=0$ is
  1. $\frac {4}{5}$
  2. $\frac {3}{4}$
  3. $\frac {1}{2}$
  4. $\frac {5}{6}$

Solution

Given equation of parabola $\begin{aligned} & x^2-2 x+3 y-2=0 \\ & \Rightarrow \quad x^2-2 x+1=-3 y+2+1 \\ & =-3 y+3 \\ & \Rightarrow \quad(x-1)^2=-3(y-1) \end{aligned}$ Distance between the vertex and focus of parabola $\begin{aligned} & =\frac{1}{4} \times \text { Length of latusrectum } \\ & =\frac{1}{4} \times 4|a|=\left|\frac{-3}{4}\right|=\frac{3}{4} \end{aligned}$

Asked in: AP EAMCET 2016

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