Given that the hyperbola has a center at (0,0), and its vertices and foci are on y-axis. This, the equation of the hyperbola is of the form
x²/a²-y²/b²=-1 (a>0, b>0)
In the equation, vertices are (0, +/-b) .
Thus,
b=60
Foci (0,+/-√(a²+b²))
thus
√(a²+60²)=65
hence solving for a²
a²=65²-60²
a²=625
a²=25²
hence the equation is:
x²/25²-y²/60²=-1
Complete the square for the given equation
x² - 2x + ____ + y² - 2y + _____ = 98
x² - 2x + (1) + y² - 2y + (1) = 98 + (1) + (1)
(x - 1)² + (x - 1)² = 100
(x - 1)² + (x - 1)² = 10²
Now the equation is in the form (x - h)² + (y - k)² = r²
Radius = 10
1 Simplify
4
.
5
+
1
.
6
9
4.5+1.69 to
6
.
1
9
6.19.
6
.
1
9
=
9
.
5
2
6.19=9.52
2 Since
6
.
1
9
=
9
.
5
2
6.19=9.52 is false, there is no solution.
No Solution
Done
The borders are shown in the picture attached.
As you can see, starting with border 1, we have 6 daises (white squares) surrounded by 10 tulips (colored squares). Through Jerry's expression we expected:
<span>8(b − 1) + 10 =
</span>8(1 − 1) + 10 =
0 + 10 =
10 tulips.
When considering border 2, we expect:
<span>8(b − 1) + 10 =
</span>8(2 − 1) + 10 =
8 + 10 =
<span>18 tulips.
Indeed, we have the 10 tulips from border 1 and 8 additional tulips, for a total of 18 tulips.
Then, consider border 3, we expect:
</span><span>8(b − 1) + 10 =
</span>8(3 − 1) + 10 =
16 + 10 =
26<span> tulips.
Again, this is correct: we have the 10 tulips used in border 1 plus other 16 tulips, for a total of 26.
Therefore, Jerry's expression is
correct.</span>