Prove that the equation $y^{2} + 2 A x + 2 B y + C = 0$ represents a parabola, whose axis is parallel to the axis of x, and find its vertex and the equation to its latus rectum.

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Solution

$y^{2} + 2 A x + 2 B y + C = 0 y^{2} + 2 B y = - 2 A x - C y^{2} + 2 B y + B^{2} = - 2 A x - C + B^{2} {(y + B)}^{2} = - 2 A (x + \frac{C}{2 A} - \frac{B^{2}}{2 A})$
Comparing with the standard equation $Y^{2} = 4 a X$
Axis of the parabola is $Y = 0$
$y + B = 0 y = - B$
Slope of the vertex $= m = 0$
Hence it is parallel to $x$ axis
Vertex of the parabola is $X = 0, Y = 0$
$x + \frac{C}{2 A} - \frac{B^{2}}{2 A} = 0, y + B = 0 \Rightarrow x = \frac{B^{2} - C}{2 A}, y = - B (\frac{B^{2} - C}{2 A}, - B)$
Equation of the latusrectum is $x = a$
Here $4 a = - 2 A \Rightarrow a = - \frac{A}{2}$
$x + \frac{C}{2 A} - \frac{B^{2}}{2 A} = - \frac{A}{2} x = \frac{B^{2}}{2 A} - \frac{C}{2 A} - \frac{A}{2} x = \frac{B^{2} - C - A^{2}}{2 A}$

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