The distance between the vertex and the focus of the parabola x2-2 x+3 y-2=0 is

Question

The distance between the vertex and the focus of the parabola x2-2 x+3 y-2=0 is (A) (4/5) (B) (3/4) (C) (1/2) (D) (5/6)

Tardigrade Admin · Accepted Answer

Equation of parabola is x2-2 x+ 3 y-2 =0 ⇒ x2-2 x+1 =-3 y+2+1 ⇒ (x-1)2 =-3(y-1) Distance between the vertex and focus of parabola =(1/4) × Length of latusrectum =(1/4) × 3=(3/4)

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

$\frac{4}{5}$

$\frac{3}{4}$

$\frac{1}{2}$

$\frac{5}{6}$

Equation of parabola is
$x^{2} - 2 x + 3 y - 2 = 0$
$\Rightarrow x^{2} - 2 x + 1 = - 3 y + 2 + 1$
$\Rightarrow (x - 1)^{2} = - 3 (y - 1)$
Distance between the vertex and focus of parabola
$= \frac{1}{4} \times$ Length of latusrectum
$= \frac{1}{4} \times 3 = \frac{3}{4}$

Q. The distance between the vertex and the focus of the parabola x2−2x+3y−2=0 is

54​

43​

21​

65​