Write the distance between the vertex and focus of the parabola $y^{2} + 6 y + 2 x + 5 = 0.$

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Solution

We have, $y^{2} + 6 y = - 2 x - 5$

$\Rightarrow y^{2} + 2 \times y \times 3 + 9 = - 2 x - 5 + 9$

$\Rightarrow (y + 3)^{2} = - 2 x + 4$

$\Rightarrow (y + 3)^{2} = - 2 (x - 2) . . . (i)$

Shifting the origin to the point (2,-3) without rotating the axes and denoting the new coordinates w.r.t. these axes by x and y.

We have, x=X+2,y=Y-3 ...(iii)

Using these relations,equation (i) reduces to $y^{2} = - 2 X$ ...(ii)

This is of the form $Y^{2} = - 4 a X,$ on comparing,we get 4a=2

$\Rightarrow a = \frac{2}{4} = \frac{1}{2}$

$∴$ The distance between the vertex and focus of the parabola is $\frac{1}{2}$ .

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List-I	List-II
A) Vertex of the parabola $y^{2} - x + 4 y + 5 = 0$	p) (1,4)
B) Focus of the parabola $x^{2} - 2 x - 12 y + 13 = 0$	q). (1,-2)
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