The equation of the directrix of parabola $y^2-x+4 y+5=0$ is

The equation of the directrix of parabola $y^2-x+4 y+5=0$ is
  1. 4y − 3 = 0
  2. 4x − 3 = 0
  3. 3x − 4 = 0
  4. 3y − 4 = 0

Solution

Given parabola is $\begin{aligned} y^2-x+4 y+5 & =0 \\ y^2+4 y & =x-5 \\ y^2+4 y+4 & =x-5+4 \\ (y+2)^2 & =(x-1)\end{aligned}$ On comparing with general form of parabola $\begin{aligned} & \qquad \quad(y-k)^2=4 a(x-h) \\ & \text { or } \quad Y^2=4 a X \\ & \text { where } Y=y-k \\ & \qquad X=x-h \\ & \text { Vertex }(h, k)=(1,-2) \\ & \text { and } \quad a=\frac{1}{4}\end{aligned}$ Since, equation of directrix is X = − a $\begin{array}{ll} & x-1=-1 / 4 \\ & x=1-1 / 4=3 / 4 \\ \Rightarrow \quad & 4 x-3=0\end{array}$

Asked in: AP EAMCET 2021 (25 Aug Shift 2)

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