Question

Describe the sequence of transformations that would take the graph of f(x)=x² to each parabola described below. Focus: (1/4,0),directrix: x=-1/4

Answer 1

Answer:

The parabola obtained is x=y²

Step-by-step explanation:

Given the focus (1/4,0) and directrix: x=-1/4

A point on the parabola is determined by the distance of that point with the focus and with the directrix:

Distance between the parabola and the directrix:$\sqrt{(x+\frac{1}{4})^{2} }$

Distance between the parabola and the focus:$\sqrt{(x-\frac{1}{4})^{2} +(y-0)^{2} }$

$\sqrt{(x+\frac{1}{4})^{2}}$=$\sqrt{(x-\frac{1}{4})^{2} +(y-0)^{2} }$

$(x+\frac{1}{4})^{2}$=$(x-\frac{1}{4})^{2} +(y-0)^{2}$

$x^2+\frac{1}{2} x+\frac{1}{16} = x^2-\frac{1}{2} x+\frac{1}{16} + y^{2}$

$\frac{1}{2} x=-\frac{1}{2} x + y^{2}$

$\frac{1}{2} x+\frac{1}{2} x = y^{2}$

x=y²