Find the vertex of the parabola $5 x^{2} + 30 x + 2 y + 59 = 0$

(3, 7)

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(- 3, - 7) .

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(3, - 5)

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(3, - 4) .

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Solution

The correct option is D $(- 3, - 7) .$
$5 x^{2} + 30 x + 2 y + 59 = 0$
$5 (x^{2} + 6 x + 9) = (- 2 y - 59) + 45 = - 2 (y + 7)$
or ${(x + 3)}^{2} = \frac{2}{5} . (y + 7)$
Above is of the form $X^{2} = 4 A Y$
where $X = x + 3, Y = y + 7, 4 A = - \frac{2}{5},$
or $A = - \frac{1}{10}$
Vertex : $X = 0, Y = 0$ or $x + 3 = 0, y + 7 = 0$
or $(- 3, - 7)$

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