AboutAbe Mantell Expertise Hello,
I am a college professor of mathematics and regularly teach all levels
from elementary mathematics through differential equations, and would
be happy to assist anyone with such questions!
Experience Over 15 years teaching at the college level.
Question 1.State the vertex of x2 – 2x – y = 0.
2.State the equation of the directrix of (y – 4)2 = –8(x – 3) .
3.State the equation of the axis of symmetry of x2 – 6x + 10y – 1 = 0.
4.State the direction of the opening of y2 + 8x – 6y – 3 = 0.
5.State the coordinates of the focus of x2 – 4y – 8 = 0
6. Write the general equation of the parabola given vertex (4, 3) and focus (6, 3).
Answer 1. y=x^2-2x has vertex at x=-b/(2a)=2/2=1, so x=1 and y=-1 ==> (1,-1)
2. Basically, if you arrange the equation for a parabola in this form:
(x-h)^2=4p(y-k)
The vertex is (h,k); the focus is (h, k+p); the directrix is y=k-p where p is the distance from the vertex to the focus.
or if of the form (y-h)^2=4p(x-k) has directrix x=k-p
You have (y-4)^2=-8(x-3) ==> (y-4)^2=4(-2)(x-3)...
so the directrix is x=3-(-2) or x=5
3. x^2-6x-1=-10y ==> y=(-1/10)x^2+(3/5)x+(1/10)
.. axis of symmetry is x=-b/(2a)=-(3/5)/(-2/10) .. x=3
4. y^2-6y-3=-8x or 8x=-y^2+6y+3 since the y^2 is negative it "opens"
.. to the left.
5. 4y=x^2-8 can be written as x^2=4y+8 or (x-0)^2=4(1)(y+4)
.. so, as explained in #2, the focus is (0,-4+1) = (0,-3)
6. (y-3)^2=4(2)(x-4), the 2 comes from 6-4...
