A hyperbola centered at the origin has vertices at (+-sqrt33,0) and foci at (+-sqrt59,0). Write the equation of this hyperbola.

Identify Hyperbola Equation: The equation of a hyperbola centered at the origin with horizontal transverse axis is given by (x^2/a^2) - (y^2/b^2) = 1, where 2a is the distance between the vertices and 2c is the distance between the foci. We are given the vertices at (±√33, 0), so a = √33.Find Distance Between Vertices and Foci: Next, we are given the foci at (±√59, 0), so c = √59. The relationship between a, b, and c in a hyperbola is c^2 = a^2 + b^2. We can use this to solve for b^2.Calculate b^2: Substitute the known values of a and c into the relationship c^2 = a^2 + b^2 to find b^2. (√59)^2 = (√33)^2 + b^2 59 = 33 + b^2 b^2 = 59 - 33 b^2 = 26Write Hyperbola Equation: Now that we have a^2 and b^2, we can write the equation of the hyperbola. Substitute a^2 = 33 and b^2 = 26 into the standard equation of the hyperbola. (x^2/33) - (y^2/26) = 1

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

A hyperbola centered at the origin has vertices at
(+-sqrt33,0) and foci at
(+-sqrt59,0).
Write the equation of this hyperbola.

A hyperbola centered at the origin has vertices at $( \pm \sqrt{33}, 0)$ and foci at $( \pm \sqrt{59}, 0)$ . $\newline$ Write the equation of this hyperbola.

View step-by-step help

Home

Math Problems

Precalculus

Convert equations of conic sections from general to standard form

Full solution

Q. A hyperbola centered at the origin has vertices at

( \pm \sqrt{33}, 0)

and foci at

( \pm \sqrt{59}, 0)

\newline

Write the equation of this hyperbola.

Identify Hyperbola Equation: The equation of a hyperbola centered at the origin with horizontal transverse axis is given by $(\frac{x^2}{a^2}) - (\frac{y^2}{b^2}) = 1$ , where $2a$ is the distance between the vertices and $2c$ is the distance between the foci. We are given the vertices at $(\pm\sqrt{33}, 0)$ , so $a = \sqrt{33}$ .
Find Distance Between Vertices and Foci: Next, we are given the foci at $(\pm\sqrt{59}, 0)$ , so $c = \sqrt{59}$ . The relationship between $a$ , $b$ , and $c$ in a hyperbola is $c^2 = a^2 + b^2$ . We can use this to solve for $b^2$ .
Calculate $b^2$ : Substitute the known values of $a$ and $c$ into the relationship $c^2 = a^2 + b^2$ to find $b^2$ . $\newline$ $(\sqrt{59})^2 = (\sqrt{33})^2 + b^2$ $\newline$ $59 = 33 + b^2$ $\newline$ $b^2 = 59 - 33$ $\newline$ $b^2 = 26$
Write Hyperbola Equation: Now that we have $a^2$ and $b^2$ , we can write the equation of the hyperbola. Substitute $a^2 = 33$ and $b^2 = 26$ into the standard equation of the hyperbola. $(\frac{x^2}{33}) - (\frac{y^2}{26}) = 1$