Check the picture below, so the hyperbola looks more or less like so, so let's find the length of the conjugate axis, or namely let's find the "b" component.
![\textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Ctextit%7Bhyperbolas%2C%20horizontal%20traverse%20axis%20%7D%20%5C%5C%5C%5C%20%5Ccfrac%7B%28x-%20h%29%5E2%7D%7B%20a%5E2%7D-%5Ccfrac%7B%28y-%20k%29%5E2%7D%7B%20b%5E2%7D%3D1%20%5Cqquad%20%5Cbegin%7Bcases%7D%20center%5C%20%28%20h%2C%20k%29%5C%5C%20vertices%5C%20%28%20h%5Cpm%20a%2C%20k%29%5C%5C%20c%3D%5Ctextit%7Bdistance%20from%7D%5C%5C%20%5Cqquad%20%5Ctextit%7Bcenter%20to%20foci%7D%5C%5C%20%5Cqquad%20%5Csqrt%7B%20a%20%5E2%20%2B%20b%20%5E2%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)

Answer:
Step-by-step explanation:
eq. of circle with center (h,k) and radius=r is
(x-h)²+(y-k)²=r²
reqd. eq. is
(x-2)²+(y+5)²=9²
Answer:
Area = 31.5 units^2
Step-by-step explanation:
7 x 3 = 21
Area of triangle is (height x base) / 2
(3 x 7) / 2 = 10.5
21 + 10.5 = 31.5
Answer:
4
Step-by-step explanation:
The equation of the line is written in the slope-intercept form, which is: y = mx + b, where m represents the slope and b represents the y-intercept. In our equation, y = − 7 x + 4 , we see that the y-intercept of the line is 4.
Add me brainiest
Answer: 59/3k+14
Step-by-step explanation: