- Thread starter
- #1

So I found the equation for the perpendicular line to the directrix in order to find the vertex, which I got the line y = x that is perpendicular to the directrix, then solved the system of equations to find the common intersection point, which was (1,1). I used the midpoint formula from the focus to the directrix, and got (0.5,0.5) (even though I know by intuition). So the vertex is at (0.5,0.5) since it is exactly in between the focus and directrix.Given directrix y = -x + 2 and focus (0,0), find the equation of the parabola

Using the form [tex](x - \frac{1}{2})^{2} = 4p(y - \frac{1}{2})[/tex], I plugged in the vertex already. To find the value of 'p' I needed to use the distance formula for between either the focus and vertex, or the vertex and directrix which I know should be exactly the same. So I did: [tex]distance = \sqrt{(\frac{1}{2} - 0)^{2} + (\frac{1}{2} - 0)^{2}}[/tex]

[tex]distance = \sqrt{(\frac{1}{4}) + (\frac{1}{4})}[/tex]

[tex]distance = \sqrt{\frac{2}{4}}[/tex]

[tex]distance = \frac{\sqrt{2}}{2}[/tex]

So:

[tex](x - \frac{1}{2})^{2} = 4(\frac{\sqrt{2}}{2})(y - \frac{1}{2})[/tex]

[tex](x - \frac{1}{2})^{2} = \frac{4\sqrt{2}}{2}(y - \frac{1}{2})[/tex]

[tex](x - \frac{1}{2})^{2} = 2\sqrt{2}(y - \frac{1}{2})[/tex]

Put in standard form:

[tex]y = \frac{x^{2} - x + \frac{4\sqrt{2} + 1}{4}}{2\sqrt{2}}[/tex]

I don't know if I've made any careless errors and I didn't want to rationalize the denominator and risk messing this up, but when I try to graph this, the graph of the parabola passes through the directrix...I thought it wasn't supposed to touch it at all?