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

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Solution

The correct option is C
2

$2 y^{2} + 3 y + 4 x - 2 = 0$
$\Rightarrow 2 (y^{2} + \frac{3}{2} y) = - 4 x + 2$
$\Rightarrow y^{2} + \frac{3}{2} y = - 2 x + 1$
$\Rightarrow y^{2} + \frac{3}{2} y = - 2 (x + \frac{1}{2})$

L.R=2

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2 y^{2} + 3 y + 4 x - 2 = 0

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List-l	List - II
A. Length of the latusrectum of the parabola $3 y^{2} = - 4 x$	1. $(5, 2)$
B.Equation of the parabola whose focus $($ 0,4 $)$ and directrix $x + 4 = 0$ is	2. $\frac{- 4}{3}$
C. The focus of the parabola $(y - 2)^{2} = 8 (x - 3)$ is	3. $\frac{4}{3}$
D. Equation of the parabola whose vertex $(- 1, - 2)$ latusrectum 4 and axis is parallel to y-axis	4. $(x + 1)^{2} = 4 (y + 2)$
	5. $x^{2} = 16 y$

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