Question

derive the equation of the parabola with a focus at (4,-7) and a directrix of y=-15. put the equatiom in standard form

Answer 1

`\sqrt{(x_(0) - 4)^(2) + (y_(0) - (-7))^(2)} = |y_(0) - (-15)|`

`(\sqrt{(x_(0) - 4)^(2) + (y_(0) + 7)^(2))^(2)} = (y_(0) + 15)^(2)`

`(x_(0) - 4)^(2) + (y_(0) + 7)^(2) = (y_(0) + 15)^(2)`

`(x_(0)^(2) - 8x_(0) + 16) + (y_(0)^(2) + 14y_(0) + 49) = y_(0)^(2) + 30y_(0) + 225`

`x_(0)^(2) + y_(0)^(2) - 8x_(0) + 14y_(0) + 16 + 49 = y_(0)^(2) + 30y_(0) + 225`

`x_(0)^(2) + y_(0)^(2) - 8x_(0) + 14y_(0) + 65 = y_(0)^(2) + 30y_(0) + 225`

`x_(0)^(2) - 8x_(0) - 16y_(0) - 160 = 0`

`16y_(0) = x_(0)^(2) - 8x_(0) - 160`

`(16y_(0))/(16) = (x_(0)^(2) - 8x_(0) - 160)/(16)`

`y_(0) = (1)/(16)x_(0)^(2) - (1)/(2)x_(0) - 10`

`y = (1)/(16)x^(2) - (1)/(2)x - 10`

Answer 2

The equation of the parabola with a focus at (4,-7) and a directrix of y=-15 is
`y=(x^(2))/(16)-(x)/(2)-10`

Explanation:

we need to drive the equation of the parabola with a focus at ( 4, -7) and a directrix of y= -15

From the given focus ( 4, -7) and equation of directrix y = - 15 calculate p

`p=(1)/(2)(y_0-y)`

where
`y_0` is is ordinate of focus and y is equation of directrix.

`p=(1)/(2)(-7-(-15))`

`p=(1)/(2)(-7+15)`

`p=(1)/(2)(8)`

`p=4`

Calculate the vertex (h,k)

`h=4\; \text{and}\; k=(-7+(-15))/(2)=-11`

vertex (h,k) =(4,-11)

Since, vertex form is :

`(x-h)^(2)=4p(y-k)` (positive 4p shows it open upward)

`(x-4)^(2)=4(4)(y-(-11))`

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

`x^(2)+16-8x=16y+176`

subtract both the sides by 176,

`x^(2)-8x-160=16y`

Divide both the sides in above by 16,

`(x^(2))/(16)-(8x)/(16)-(160)/(16)=y`

`(x^(2))/(16)-(x)/(2)-10=y`

Hence, the equation of the parabola with a focus at (4,-7) and a directrix of y=-15 is
`y=(x^(2))/(16)-(x)/(2)-10`