Question

Find the equation of a parabola with vertex at the origin and the directrix,y=2.

Answer 1

$Let(x_1,y_1)\text{be the coordinates of the point intersection of the axis and the directrix}\therefore (x_1,y_1)=(0,2)[\because y=2]\text{We know that,the vertex is the mid-point of the line segment joining (0,2)}andfocus(x_2,y_2)\therefore \frac{x_2+0}{2}and\frac{y_2+2}{2}=0[\because vertexattheorigin]\therefore x_2=0and{y}_2=-2\therefore \text{The coordinates of focus is (0,-2)}.\text{By the definition of parabola PS=PM}\Rightarrow P{S}^2=P{M}^2\Rightarrow (x-0{)}^2+(y+2{)}^2={\left[\frac{y-2}{\sqrt{1}}\right]}^2\Rightarrow x^2+y^2+4+4y=(y-2{)}^2\Rightarrow x^2+y^2+4+4y=y^2+4-4y\Rightarrow x^2=-4y-4y\Rightarrow x^2=-8y\text{Hence,the required equation of parabola is}{x}^2=-8y$