Question

What is the equation of the parabola with a focus at (8,2) and a directrix of y= 5?

Answer 1

The equation is `(x-8)^2=-3(2y-7)`

Explanation:

Any point on the parabola is equidistant from the focus and the directrix

Therefore,

`sqrt((x-8)+(y-2))=5-y`

Squaring,

`(x-8)^2+(y-2)^2=(5-y)^2`

`(x-8)^2+cancely^2-4y+4=25-10y+cancely^2`

`(x-8)^2=-6y+21`

`(x-8)^2=-3(2y-7)`

graph{((x-8)^2+3(2y-7))(y-5)((x-8)^2+(y-2)^2-0.1)=0 [-32.47, 32.47, -16.24, 16.25]}

Answer 2

`x^2-16x+6y+43=0`

Explanation:

`"for any point "(x,y)" on the parabola"`

`"the distance from "(x,y)" to the focus and directrix"`
`"are equal"`

`"using the "color(blue)"distance formula"" and equating"`

`rArrsqrt((x-8)^2+(y-2)^2)=|y-5|`

`color(blue)"squaring both sides"`

`(x-8)^2+(y-2)^2=(y-5)^2`

`rArrx^2-16x+64+y^2-4y+4=y^2-10y+25`

`rArrx^2-16x+64cancel(+y^2)-4y+4cancel(-y^2)+10y-25=0`

`rArrx^2-16x+6y+43=0`