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

To find the quadratic equation:

Assume any point (x, y) on parabola.

Use the distance formula i.e,
`√((x_1-x_2)^2+(y_1-y_2)^2)`

Distance between (x , y) and focus
`(-4 , (-5)/(4))` is:

`\sqrt{(x+4)^2 +(y+(5)/(4))^2 }`

Now, the distance between (x, y) and directrix of y =
`(27)/(4)` is:

`\sqrt{(y-(27)/(4))^2 }`

On the parabola these distance are same;

`\sqrt{(x+4)^2 +(y+(5)/(4))^2 }` =
`\sqrt{(y-(27)/(4))^2 }`

`(x+4)^2 + (y+(5)/(4))^2 = (y-(27)/(4))^2`

`(x+4)^2 + y^2+(25)/(16)+(5)/(2)y = y^2+(729)/(16)-(27)/(2)y`

or

`(x+4)^2 = y^2+(729)/(16)-(27)/(2)y -y^2-(25)/(16)-(5)/(2)y`

Simplify:

`(x+4)^2 = (729-25)/(16)-(27+5)/(2)y`

`(x+4)^2 = (704)/(16)-(32)/(2)y`

Simplify:

`(x+4)^2 =44-16y`

or

`16y = -(x+4)^2+44`

Divide both sides by 16 we get;

`y = (-1)/(16)(x+4)^2 + (44)/(16)`

`y = (-1)/(16)(x^2+16+8x) + (11)/(4)`

or

`y= (-1)/(16)x^2-1-(1)/(2)x+ (11)/(4)`

Simplify:

`y= (-1)/(16)x^2-(1)/(2)x+ (7)/(4)`

therefore, the equation of the quadratic is; f(x) =negative one sixteenthsx^2 − one half x + seven-fourths.