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A hyperbola centered at the origin has vertices at 
(+-sqrt7,0) and foci at 
(+-sqrt27,0).
Write the equation of this hyperbola.

A hyperbola centered at the origin has vertices at (±7,0) ( \pm \sqrt{7}, 0) and foci at (±27,0) ( \pm \sqrt{27}, 0) .\newlineWrite the equation of this hyperbola.

Full solution

Q. A hyperbola centered at the origin has vertices at (±7,0) ( \pm \sqrt{7}, 0) and foci at (±27,0) ( \pm \sqrt{27}, 0) .\newlineWrite the equation of this hyperbola.
  1. Hyperbola Equation: The standard form of the equation of a hyperbola centered at the origin with horizontal transverse axis is (x2/a2)(y2/b2)=1(x^2/a^2) - (y^2/b^2) = 1, where 2a2a is the distance between the vertices and 2c2c is the distance between the foci. We are given the vertices at (±7,0)(\pm\sqrt{7}, 0), which means a=7a = \sqrt{7}. We are also given the foci at (±27,0)(\pm\sqrt{27}, 0), which means c=27c = \sqrt{27}.
  2. Finding Value of b: We need to find the value of bb. The relationship between aa, bb, and cc in a hyperbola is c2=a2+b2c^2 = a^2 + b^2. We already know that a=7a = \sqrt{7} and c=27c = \sqrt{27}. Let's plug these values into the equation to find b2b^2.\newlinec2=a2+b2c^2 = a^2 + b^2\newline(27)2=(7)2+b2(\sqrt{27})^2 = (\sqrt{7})^2 + b^2\newlineaa00\newlineaa11\newlineaa22
  3. Writing Hyperbola Equation: Now that we have b2=20b^2 = 20, we can write the equation of the hyperbola. The equation is (x2a2)(y2b2)=1(\frac{x^2}{a^2}) - (\frac{y^2}{b^2}) = 1. Substituting a2=7a^2 = 7 and b2=20b^2 = 20 into the equation, we get:(x27)(y220)=1(\frac{x^2}{7}) - (\frac{y^2}{20}) = 1

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