Question

Find the coordinates of vertices and foci and the equation of the directrices and eccentricity and axes of the conic x^2-8x+2y+7=0

Answer 1

x^2-8x+2y+7=0
x^2-8x+16+2y+7-16=0
(x-4)^2+2y-9=0
(x-4)^2= -2(y-9/2)
it is an equation of parabola...
X^2=4AY
On comparing...
A=-1/2
now vertice of parabola is (4,9/2)
foci is (0,-1/2)
equation of directrix is y=1/2
axes of the conic is y-axis
eccentricity is 1

hope this helps:p

Answer 2

x^2 - 8x +2y+7=0
(x-4)^2 -16 +2y+7=0
(x-4)^2=-2(y-9/2)
it is an parabolic equation of the form
X^2=-4BY
its eccentricity=1
on comparing
B=1/2
vertices V(4,9/2)
focus (4,4)
equation of directrix y=5
equation of the conic x=4