Equation of the parabola whose vertex is (−3,−2), axis is horizontal and which passes through the point (1,2) is
A
y2+4y+4x−8=0
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B
y2+4y−4x+8=0
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C
y2+4y−4x−8=0
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D
none of these
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Solution
The correct option is Cy2+4y−4x−8=0 Since the axis is horizontal and vertex is (−3,−2) and passes through (1,2). ∴ the equation of the parabola must be of the form (y+2)2=4a(x+3)
It passes through (1,2), ∴16=16a⇒a=1
Hence, the equation of the required parabola is (y+2)2=4(x+3)
or y2+4y−4x−8=0