Find an equation whose graph is a parabola with directrix y= -1 and vertex (1,3)?

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Answer 1 · 2017-06-21T11:32:36

The equation of parabola is $y = 1/16 (x-1)^2+3$

The equation of parabola in vertex form is $y=a(x-h)^2+k ; (h,k)$ being vertex.

The equation of parabola is $y = a (x-1)^2+3$

The vertex is $(1,3)$ and directrix is $y=-1$ .

The distance between vertex and directrix is $d=|3-(-1)| =4$ . The directrix is below the vertex, so parabola opens upwards and $a>0$

We know $d= 1/(|4a|) or a = 1/(4d)= 1/(4*4) =1/16$

So the equation of parabola is $y = 1/16 (x-1)^2+3$

graph{1/16(x-1)^2+3 [-40, 40, -20, 20]} [Ans]