Find the latus rectum of the parabola $x^{2} + 2 y - 3 x + 5 = 0$

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Solution

The correct option is B $2$
$x^{2} + 2 y - 3 x + 5 = 0$
$\Rightarrow x^{2} - 3 x = - 2 y - 5$
$\Rightarrow x^{2} - 2. \frac{3}{2} x + \frac{9}{4} = - 2 y - 5 + \frac{9}{4} = - 2 y - \frac{11}{4}$
$\Rightarrow {(x - \frac{3}{2})}^{2} = - 2 (y - \frac{11}{4})$
Hence latus rectum $= 4 \times \frac{1}{2} = 2$

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