# Find the equation of the parabola with focus (2;1) and vertex in the origin.

I have to solve a problem which says to find the equation of the parabola with focus in A(2;1) and vertex in the origin.

Any suggestions are welcome.

• Parabola is a locus of points equidistant from the focus and a directrix. Having the vertex you can find the directrix, the rest is quite straightforward. – CiaPan Sep 10 '15 at 10:57

The axis of the parabola is the straight line passing through the focus $F=(2,1)$ and the vertex $V=(0,0)$. So the equation of the axis $h$ is $y=\dfrac{1}{2}x$.
The directrix is a straight line $d$ orthogonal to the axis and such that its distance from the vertex $Vd$ is the same as the distance between the vertex and the focus $VF$. So the slope of the directrix is $m=-2$ and it pass thorough $P=(-2,-1)$ so we can find its equation that is $2x+y+5=0$.
Now the parabola is the locus of point $X=(x,y)$ such that the distance from the directrix is the same as the distance from the focus, and writing this we find the equation: $$\dfrac{|2x+y+5|}{\sqrt{5}}=\sqrt{(x-2)^2+(y-1)^2}$$ Squaring you have the equation of the parabola.
HINT...the focus $F$ and vertex $O$ lie on the axis of symmetry. The directrix intersects the axis of symmetry at a point $K$ which is such that the vertex is the midpoint of $FK$. The directix is the line through $K$ perpendicular to the line of symmetry.
Having found the equation of the directrix, use the formula for the perpendicular distance from a point to a line to write down the requirement that the variable point $P(x,y)$ is equidistant from the focus and from the directrix.