Equation of the parabola whose vertex is $(- 3, - 2)$ , axis is horizontal and which passes through the point $(1, 2)$ is

y^{2} + 4 y + 4 x - 8 = 0

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y^{2} + 4 y - 4 x + 8 = 0

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y^{2} + 4 y - 4 x - 8 = 0

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Solution

The correct option is B $y^{2} + 4 y - 4 x - 8 = 0$
Since the axis is horizontal and vertex is $(- 3, - 2)$
$∴$ the equation of the parabola must be of the form ${(y + 2)}^{2} = 4 a (x + 3)$
It passes through $(1, 2)$ , so $16 = 16 a$
$\Rightarrow a = 1$
Hence, the equation of the parabola is
${(y + 2)}^{2} = 4 (x + 3)$
$\Rightarrow y^{2} + 4 y - 4 x - 8 = 0$

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Equation of the parabola whose vertex is (−3,−2), axis is horizontal and which passes through the point (1,2) is

The correct option is B y2+4y−4x−8=0Since the axis is horizontal and vertex is (−3,−2)∴ the equation of the parabola must be of the form (y+2)2=4a(x+3)It passes through (1,2), so 16=16a⇒a=1Hence, the equation of the parabola is(y+2)2=4(x+3)⇒y2+4y−4x−8=0

Equation of the parabola whose vertex is $(- 3, - 2)$ , axis is horizontal and which passes through the point $(1, 2)$ is