The equation of the latus rectum of the parab | vedclass

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Question

(C) Given the equation of the parabola: $x^2 + 4x + 2y = 0$. Completing the square for $x$: $(x^2 + 4x + 4) = -2y + 4$. $(x + 2)^2 = -2(y - 2)$. Compa

Vedclass · Accepted Answer

$2y = 3$

Vedclass · Answer

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

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

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