The parabola x 2 = 8y is shifted right 1 and down 7. Determine the equation for the new parabola. Then find the vertex, focus, and directrix of the new parabola. x 2 = 8y
Comparing with parabola equation x 2 = 4ay, we get a = 2.
For parabola x2=4ay
Vertex: (0,0)
Focus: (0,a)
Directrix: y = -a
Hence, for parabola x2 = 8y and a =2 and we have
Vertex : (0,0)
Focus: (0,2)
Directrix: y = -2
We are shifting parabola to right by 1 and down by 7. Hence, parabola equation will be
8(y+7)=(x-1) 2
Parabola axis becomes x = 1 as parabola is shifted to right by 1.
Vertex: (1,-7)
Focus will remain on axis; hence, x-coordinate will be 1. The distance between vertex and focus is 2. Hence, y-coordinate will be -5 and focus will be (1,-5).
The distance between directrix and vertex will be to the opposite of focus. Hence, ditrectrix will be y = -9.
The hyperbola x 2
4
y 2
5
 1 is shifted right 2 and up 2. Determine the equation for the new hyperbola. Then find the center, foci, vertices, and asymptotes of the new hyperbola.
In the given hyperbola, center is (0,0). It is shifted to right by 2 and up by 2. Hence, hyperbola equation will be
(x-2) 2 /4-(y-2) 2 /5=1.
Center will bt at (2,2).
Eccentricity = sqrt(1+b 2 /a 2 ) = sqrt(1+25/16)=
√41/4
Hence, focus of the given hyperbola is at( √41.0) as a = 4.
Hence, focus of shiftef hyperbola will be at (2+ √41,0).
Similarly directricx of given hyperbola is x = 4/√41.
Hence, directriof shiftef hyperbola will be x = 2- 4/√41
Slope of asymptotes = ±5/4. The asymptotes pass through(2,2). Hence, the equations will be y-2=(5/4)(x-2)
Or, y = (5/4)x-1/2
Other asymptote will be y-2=-(5/4)(x-2)

Or, y = -(5/4)x-9/2
Find the equation of the ellipse centered at the origin with eccentricity, e = 0.2, and foci at (

8, 0).
Focus is (ae,0). Hence, ae = 8 or, a = 8/e = 8/0.2 = 40
Eccentricity e = sqrt(1-b 2 /a 2 )
0.2=sqrt(1- b 2 /a 2 )
Or, b 2 /a 2 = 1-0.04 = 0.96
Or, b 2 = 1600*0.96 = 1536
Hence, ellipse equation is given by x 2 /1600+y 2 /1536 = 1