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 $= \left(\frac{9}{4}-1, -1\right) = \left(\frac{5}{4}, -1\right)$.

List-$A$	List-$B$
$(A)$. The vertex of the parabola $y^2+4x-2y+3=0$ is	$(I)$. $\left(\frac{5}{4}, 1\right)$
$(B)$. The vertex of the parabola $x^2+8x+12y+4=0$ is	$(II)$. $\left(1, \frac{5}{4}\right)$
$(C)$. The focus of the parabola $y^2-x-2y+2=0$ is	$(III)$. $\left(-\frac{1}{2}, 1\right)$
$(D)$. The focus of the parabola $x^2-2x-8y-23=0$ is	$(IV)$. $(1, -1)$
	$(V)$. $(-4, 1)$

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

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