Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Explore thousands of questions and answers from a knowledgeable community of experts ready to help you find solutions. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
An equation in standard form for a hyperbola with center (0, 0), vertex (-5, 0), and focus (-6, 0) is given by y²/25 - x²/9 = 1.
What is an equation?
An equation can be defined as a mathematical expression which is used to show and indicate that two (2) or more numerical quantities are equal.
How to determine the equation of a hyperbola?
Mathematically, the equation of a hyperbola in standard form is given by:
[tex]\frac{(y\;-\;k)^2}{a^2} - \frac{(x\;-\;h)^2}{b^2} = 1[/tex]
Given the following data:
Center (h, k) = (0, 0)
Vertex (h+a, k) = (-5, 0)
Foci, F = (h+c, k) = (-6, 0) and F' = (6, 0)
Also, we can logically deduce that the value of a and c are -5 and -6 respectively.
For the value of b, we would apply Pythagorean's theorem:
c² = a² + b²
b² = c² - a²
b² = (-6)² - (-5)²
b² = 36 - 25
b² = 9.
b = √9
b = 3.
Substituting the given parameters into the equation of a hyperbola in standard form, we have;
[tex]\frac{(y\;-\;k)^2}{a^2} - \frac{(x\;-\;h)^2}{b^2} = 1\\\\\frac{(y\;-\;0)^2}{-5^2} - \frac{(x\;-\;0)^2}{3^2} = 1\\\\\frac{y^2}{-5^2} - \frac{x^2}{3^2} = 1\\\\\frac{y^2}{25} - \frac{x^2}{9} = 1[/tex]
y²/25 - x²/9 = 1.
In conclusion, we can logically deduce that an equation in standard form for a hyperbola with center (0, 0), vertex (-5, 0), and focus (-6, 0) is given by y²/25 - x²/9 = 1.
Read more on hyperbola here: https://brainly.com/question/3405939
#SPJ1
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.