Answer:
Step-by-step explanation:
Hello,
Vertices are on a line parallele at ox (y=-3)
The hyperbola is horizontal.
Equation is (x-h)²/a²- (y-k)²/b²=1
Center =middle of the vertices=((-2+6)/2,-3)=(2,-3)
(h+a,k) = (6,-3)
(h-a,k)=(-2,-3)
==>k=-3 and 2h=4 ==>h=2
==>a=6-h=6-2=4 (semi-transverse axis)
Foci: (h+c,k) ,(h-c,k)
h=2 ==>c=8-2=6
c²=a²+b²==>b²=36-4²=20
Equation is:
Answer:
1
Step-by-step explanation:
First, convert all the secants and cosecants to cosine and sine, respectively. Recall that and .
Thus:
Let's do the first part first: (Recall how to divide fractions)
For the second term:
So, all together: (same denominator; combine terms)
Note the numerator; it can be derived from the Pythagorean Identity:
Thus, we can substitute the numerator:
Everything simplifies to 1.
Answer:
x = 2
Step-by-step explanation:
The product of distances from the intersection of secants to the near and far intersections with the circle are the same. For a tangent, the near and far points of intersection with the circle are the same. This relation tells us ...
(2√3)(2√3) = x(x +4)
12 = x² +4x
16 = x² +4x +4 . . . . . add the square of half the x-coefficient to complete the square
4² = (x +2)² . . . . . . . . write as squares
4 = x +2 . . . . . . . . . . positive square root
2 = x . . . . . . . . . . . . . subtract 2
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<em>Alternate solution</em>
If you believe x to be an integer, you can look for factors of 12 that differ by 4.
12 = 1×12 = 2×6 = 3×4
The factors 2 and 6 differ by 4, so x=2 and x+4=6.
Answer:
hope its help u
Step-by-step explanation: