The ends of the latus rectum of the parabola | vedclass

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(C) The given equation of the parabola is $x^2 = -8y$.Comparing this with the standard form $x^2 = -4ay$,we get $4a = 8$,which implies $a = 2$.The foc

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$(-4, -2)$ and $(4, -2)$

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(C) The given equation of the parabola is $x^2 = -8y$.Comparing this with the standard form $x^2 = -4ay$,we get $4a = 8$,which implies $a = 2$.The focus of the parabola $x^2 = -4ay$ is $(0, -a)$,so the focus is $(0, -2)$.The latus rectum is a line parallel to the $x$-axis passing through the focus $(0, -2)$,given by the equation $y = -2$.To find the ends of the latus rectum,substitute $y = -2$ into the parabola equation $x^2 = -8y$:$x^2 = -8(-2) = 16$$x = \pm 4$.Thus,the ends of the latus rectum are $(4, -2)$ and $(-4, -2)$.

The ends of the latus rectum of the parabola $x^2 + 8y = 0$ are

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