The point of intersection of the latus rectum | vedclass

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(B) The point of intersection of the latus rectum and the axis of a parabola is its focus. Given equation: $y^2+4x+2y-8=0$. Rearranging the terms: $y^

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$\left(\frac{5}{4}, -1\right)$

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(B) The point of intersection of the latus rectum and the axis of a parabola is its focus. Given equation: $y^2+4x+2y-8=0$. Rearranging the terms: $y^2+2y = -4x+8$. Completing the square for $y$: $y^2+2y+1 = -4x+8+1$. $(y+1)^2 = -4x+9$. $(y+1)^2 = -4\left(x-\frac{9}{4}\right)$. Comparing this with the standard form $(y-k)^2 = 4a(x-h)$,we get $h = \frac{9}{4}$,$k = -1$,and $4a = -4$,which implies $a = -1$. The focus is given by $(h+a, k)$. Focus $= \left(\frac{9}{4}-1, -1\right) = \left(\frac{5}{4}, -1\right)$.

The point of intersection of the latus rectum and axis of the parabola $y^2+4x+2y-8=0$ is

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