The equation of the directrix of the parabola $3 x^{2}=16 y$ is

The equation of the directrix of the parabola $3 x^{2}=16 y$ is
  1. $3 y+4=0$
  2. $3 x+4=0$
  3. $3 y-4=0$
  4. $3 x-4=0$

Solution

$3 x^{2}=16 y$ $x^{2}=\frac{16}{3} \quad ; a=\frac{4}{3}$ Directrix $\rightarrow$ $y+\frac{4}{3}=0$ $3 y+4=0$

Asked in: MHT CET 2020 (19 Oct Shift 2)

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