Question

What is the equation of the quadratic graph with a focus of (−4, −five fourths) and a directrix of y = twenty seven fourths?

Answer 1

f(x) =negative one sixteenthsx2 − one half x +seven-fourths  

Answer 2

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.