The equation of the latus rectum of the parab | vedclass

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(C) Given equation: ${x^2} + 4x + 2y = 0$Completing the square for $x$: ${x^2} + 4x = -2y$${x^2} + 4x + 4 = -2y + 4$${(x + 2)^2} = -2(y - 2)$Comparing

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$2y = 3$

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(C) Given equation: ${x^2} + 4x + 2y = 0$Completing the square for $x$: ${x^2} + 4x = -2y$${x^2} + 4x + 4 = -2y + 4$${(x + 2)^2} = -2(y - 2)$Comparing this with the standard form ${(x - h)^2} = 4a(y - k)$,we get $4a = -2$,so $a = -\frac{1}{2}$.The vertex is $(h, k) = (-2, 2)$.The equation of the latus rectum for a downward parabola is $y = k + a$.$y = 2 + (-\frac{1}{2}) = \frac{3}{2}$.Therefore,$2y = 3$.

The equation of the latus rectum of the parabola ${x^2} + 4x + 2y = 0$ is

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