But how do you get there? Complete the square to determine whether the equation represents an
ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, find the center,
foci, vertices, and lengths of the major and minor axes. If it is a parabola, find the vertex, focus,
and directrix. If it is a hyperbola, find the center, foci, vertices, and asymptotes.
Solution
y^2 = 4x + 8y
so, y^2 - 8y = 4x
y^2 - 8y + 16 = 4x +16
(y-4)^2 = 4(x+4), now we can clearly see that it is Parbola written in Form (y - y0)^2 = 4*a* (x-
x0)
where (xo, y0) is vetrex
so in our case = (x0,y0) = ( -4,4)
4*a = 4, a = 1
So focus can easily be determined as it is point on line y = 4 at distance 1 from vertex on right
side
so F = (-3,4)
now directrix can also be determind similarly as it is perpendicular to axis at distance = 1 from
vertex
so directrix , x = -5
So, It is parabola with vertex = (-4,4)
Focus = (-3,4)
Directrix , x= -5.
But how do you get there Complete the square to determine whether t.pdf
1. But how do you get there? Complete the square to determine whether the equation represents an
ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, find the center,
foci, vertices, and lengths of the major and minor axes. If it is a parabola, find the vertex, focus,
and directrix. If it is a hyperbola, find the center, foci, vertices, and asymptotes.
Solution
y^2 = 4x + 8y
so, y^2 - 8y = 4x
y^2 - 8y + 16 = 4x +16
(y-4)^2 = 4(x+4), now we can clearly see that it is Parbola written in Form (y - y0)^2 = 4*a* (x-
x0)
where (xo, y0) is vetrex
so in our case = (x0,y0) = ( -4,4)
4*a = 4, a = 1
So focus can easily be determined as it is point on line y = 4 at distance 1 from vertex on right
side
so F = (-3,4)
now directrix can also be determind similarly as it is perpendicular to axis at distance = 1 from
vertex
so directrix , x = -5
So, It is parabola with vertex = (-4,4)
Focus = (-3,4)
Directrix , x= -5
