What will be the equation of the latus rectum of parabola $y^{2} - 2 x - 6 y + 5 = 0$ ?

2 x + 3 = 0

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2 x + 1 = 0

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2 x - 3 = 0

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2 x - 1 = 0

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Solution

The correct option is A $2 x + 3 = 0$
Given the equation of parabola $y^{2} - 2 x - 6 y + 5 = 0$
On rearranging the equation we get,
$\Rightarrow y^{2} - 6 y + 5 = - 2 x$
$\Rightarrow y^{2} - 6 y + 9 - 4 = - 2 x$
$\Rightarrow (y - 3)^{2} = - 2 x + 4$
$\Rightarrow (y - 3)^{2} = - 2 (x - 2)$

From above it was concluded that the equation of parabola is of the form
$(y - h)^{2} = - 4 a (x - h)$ .

Vertex of the translated parabola is
$(2, 3) and a = \frac{1}{2} .$
So, the equation of latus rectum of translated parabola is given by $x = - a + h$ i.e.
$x = - \frac{1}{2} + 2 \Rightarrow 2 x = - 3.$
Hence, the equation of latus rectum is $2 x + 3 = 0.$

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