The equation of the parabola with the focus $(3,0)$ and the directrix $x+3=0$, is.

The equation of the parabola with the focus $(3,0)$ and the directrix $x+3=0$, is.
  1. $y^2=3 x$
  2. $y^2=6 x$
  3. $y^2=12 x$
  4. $y^2=2 x$

Solution

Given that focus is $S(3,0)$, let $P(x, y)$ be any point on the parabola. $\therefore$ Directrix is, $x+3=0$ Also, $S P^2=P M^2$ $ \begin{aligned} & \Rightarrow \quad(x-3)^2+(y-0)^2=\left(\frac{x+3}{\sqrt{1}}\right)^2 \\ & \Rightarrow \quad(x-3)^2+y^2=(x+3)^2 \\ & \Rightarrow \quad y^2=(x+3)^2-(x-3)^2 \\ & \Rightarrow \quad=(x+3+x-3)(x+3-x+3) \\ & \Rightarrow \quad y^2=2 x 6=12 x \end{aligned} $

Asked in: AP EAMCET 2002

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