# Find an equation in standard form for the conic. Calculus II

1. May 7, 2012

### StudentofSci

1. The problem statement, all variables and given/known data

Find an equation in standard form for the conic ,
a)Ellipse: vertices (0,0) & (0,8), foci (0,2) & (0,6).
b) hyperbola: vertices (-3,0) & (3,0), asymptotes y= ±2x

2. Relevant equations
Ellipse: (x-h)^2/b^2+(y-k)^2/a^2=1, vertices ± a from center, foci ± c from center, c^2=a^2-b^2, a<b, ceneter = (h,k)
Hyperbola: (x-h)^2/a^2-(y-k)^2/b^2=1, vertices ± a from center, foci ± c from center, c^2=a^2+b^2, ceneter = (h,k), asymptotes= k±b/a(x-h)

3. The attempt at a solution

a) foci are ±c from center, the foci are (0,2) & (0,6) thus h must = 0, c must= 2, c^2=4, and k= 4
I know c, and the center (0,4)
vertices are ± a from the center thus if the center is (0,4) when vertices are ( 0,0) & (0,8), a must= 4 and a^2=16
since I know have a^2 and c^2 I can solve for b^2
4=16-b^2, b^2=12
I can now solve the problem and write the equation in the form of (x-h)^2/b^2+(y-k)^2/a^2=1.
x^2/16+(y-4)^2/16=1
End part A

b)
vertices are ± a from center @ (±3,0) thus the center is (0,0) h=0, k=0 and a=3 and a^2=9
asymptotes are in the form of k±b/a(x-h) thus 0 ± b/3(x-0)= ±2x, b must =6 b^2=36
I can now write the equation in the form of (x-h)^2/a^2-(y-k)^2/b^2=1 as:
x^2/9-y^2/36=1

End part B

That is all my work/solutions for this problem.
Any help with what I may have done wrong and how to learn how to correct is, or even just saying "correct" is appreciated thank you.

2. May 7, 2012

### HallsofIvy

Staff Emeritus
Assuming an ellipse is set up with foci on the x-axis, we can take them to be at (c, 0) and (-c, 0). We can take the vertices on the major axis to be at (a, 0) and (-a, 0). The distance from the focus (-c, 0) to (a, 0) is a-(-c)= a+ c and the distance from that vertex to the other focus, (c, 0) is a- c. That is, the total distance from focus to vertex to other vertex, along the major axis, is a+ c+ a- c= 2a.

The distance from the focus (c, 0) to the vertex on the minor axis, (0, b) is $\sqrt{b^2+ c^2}$. The distance from (0, b) to (-c, 0) is also $\sqrt{b^2+ c^2}$ so the distance form focus to vertex to the other vertex, on the minor axis is $2\sqrt{b^2+ c^2}$.

Now, a defining property of an ellipse is that the distance from one focus, to any point on the ellipse, back to the other vertex is a constant. Therefore, we must have $2\sqrt{b^2+ c^2}= 2a$ so, dividing both sides by 2 and squaring both sides, $b^2+ c^2= a^2$. I would be surprised if you were not given that formula in you class or textbook.

You are given a and c so you can use that to find b.