Answer:
To find the quadratic equation:
Assume any point (x, y) on parabola.
Use the distance formula i.e, \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
Distance between (x , y) and focus \((-4 , \frac{-5}{4})\) is:
\(\sqrt{(x+4)^2 +(y+\frac{5}{4})^2 }\)
Now, the distance between (x, y) and directrix of y = \(\frac{27}{4}\) is:
\(\sqrt{(y-\frac{27}{4})^2 }\)
On the parabola these distance are same;
\(\sqrt{(x+4)^2 +(y+\frac{5}{4})^2 }\) = \(\sqrt{(y-\frac{27}{4})^2 }\)
\((x+4)^2 + (y+\frac{5}{4})^2 = (y-\frac{27}{4})^2\)
\((x+4)^2 + y^2+\frac{25}{16}+\frac{5}{2}y = y^2+\frac{729}{16}-\frac{27}{2}y\)
or
\((x+4)^2 = y^2+\frac{729}{16}-\frac{27}{2}y -y^2-\frac{25}{16}-\frac{5}{2}y\)
Simplify:
\((x+4)^2 = \frac{729-25}{16}-\frac{27+5}{2}y\)
\((x+4)^2 = \frac{704}{16}-\frac{32}{2}y\)
Simplify:
\((x+4)^2 =44-16y\)
or
\(16y = -(x+4)^2+44\)
Divide both sides by 16 we get;
\(y = \frac{-1}{16}(x+4)^2 + \frac{44}{16}\)
\(y = \frac{-1}{16}(x^2+16+8x) + \frac{11}{4}\)
or
\(y= \frac{-1}{16}x^2-1-\frac{1}{2}x+ \frac{11}{4}\)
Simplify:
\(y= \frac{-1}{16}x^2-\frac{1}{2}x+ \frac{7}{4}\)
therefore, the equation of the quadratic is; f(x) =negative one sixteenthsx^2 − one half x + seven-fourths.
