Question

What is the equation for a parabola with vertex:(8,6) and focus:(3,6)?

Answer 1

For the parabola it is given

`V->"Vertex"=(8,6)`

`F->"Focus"=(3,6)`

We are to find out the equation of the parabola

The ordinates of V(8,6) and F(3,6) being 6 the axis of parabola will be parallel to x-axis and its equation is `y=6`

Now let the coordinate of the point (M) of intersection of directrix and axis of parabola be `(x_1,6)`.Then V will be mid point of MF by the property of parabola. So

`(x_1+3)/2=8=>x_1=13`
`"Hence"M->(13,6)`

The directrix which is perpendicular to the axis (`y=6`) will have equation `x=13 or x-13=0`

Now if`P(h,k)` be any point on the parabola and N is the foot of the perpendicular drawn from P to the directrix,then by the property of parabola

`FP=PN`

`=>sqrt((h-3)^2+(k-6)^2)=h-13`

`=>(h-3)^2+(k-6)^2=(h-13)^2`

`=>(k-6)^2=(h-13)^2-(h-3)^2`

`=>(k^2-12k+36=(h-13+h-3)(h-13-h+3)`

`=>k^2-12k+36=(2h-16)(-10)`

`=>k^2-12k+36+20h-160=0`

`=>k^2-12k+20h-124=0`

Replacing h by x and k by y we get the required equation of the parabola as

`color(red)(y^2-12y+20x-124=0)`