Answered

Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Get immediate answers to your questions from a wide network of experienced professionals on our Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

A hyperbola centered at the origin has verticies at (add or subtract square root of 61,0 and foci at (add or subtract square root of 98,0

Sagot :

raphealnwobi

Answer:

[tex]\frac{x^2}{61}-\frac{y^2}{37} =1[/tex]

Step-by-step explanation:

The standard equation of a hyperbola is given by:

[tex]\frac{(x-h)^2}{a^2} -\frac{(y-k)^2}{b^2} =1[/tex]

where (h, k) is the center, the vertex is at (h ± a, k), the foci is at (h ± c, k) and c² = a² + b²

Since the hyperbola is centered at the origin, hence (h, k) = (0, 0)

The vertices is (h ± a, k) = (±√61, 0). Therefore a = √61

The foci is (h ± c, k) = (±√98, 0). Therefore c = √98

Hence:

c² = a² + b²

(√98)² = (√61)² + b²

98 = 61 + b²

b² = 37

b = √37

Hence the equation of the hyperbola is:

[tex]\frac{x^2}{61}-\frac{y^2}{37} =1[/tex]

We appreciate your time. Please come back anytime for the latest information and answers to your questions. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.