Question 1121260: Find the Standard equation of hyperbola, center, foci, vertices at asymptotes of the function 4x²-5y²+32x+30y=1.
Thank you! Found 2 solutions by solver91311, ikleyn:Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website! .
Find the Standard equation of hyperbola, center, foci, vertices at asymptotes of the function 4x² - 5y² + 32x + 30y = 1.
Thank you!
~~~~~~~~~~~~~~
4x^2 - 5y^2 + 32x + 30y = 1 ====> Complete the squares for x- and y-terms separately ====>
(4x^2 + 32x) + (-5y^2 + 30y) = 1
4(x^2 + 8x) - 5*(y^2 - 6y) = 1
4*(x^2 + 8x + 16) - 5*(y^2 - 6y + 9) = 1 + 4*16 - 5*9
4*(x+4)^2 - 5*(y-3)^2 = 20. <<<---=== Divide by 20 both sides
- = 1
- = 1
It is the standard form equation for a hyperbola.
The hyperbola has the center at (x,y) = (-4,3).
Real axis is parallel to x-axis; imaginary axis is parallel to y-axis.
Real semi-axis is units long; Imaginary axis is 2 units long.
Vertices are at y= 3: x = and x= .
The foci are x= = -4 - 3 = -7 and x= = -4 + 3 = -1.
The referred lesson is the part of this online textbook under the topic
"Conic sections: Hyperbolas. Definition, major elements and properties. Solved problems".
Save the link to this textbook together with its description
Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson