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balandron [24]
3 years ago
11

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

Mathematics
2 answers:
MariettaO [177]3 years ago
4 0

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.


horrorfan [7]3 years ago
4 0
<span> f(x) =negative one sixteenthsx2 − one half x +seven-fourths  </span>
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