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A hyperbola centered at the origin has vertices at 
(0,+-sqrt12) and foci at 
(0,+-sqrt45).
Write the equation of this hyperbola.

A hyperbola centered at the origin has vertices at (0,±12) (0, \pm \sqrt{12}) and foci at (0,±45) (0, \pm \sqrt{45}) .\newlineWrite the equation of this hyperbola.

Full solution

Q. A hyperbola centered at the origin has vertices at (0,±12) (0, \pm \sqrt{12}) and foci at (0,±45) (0, \pm \sqrt{45}) .\newlineWrite the equation of this hyperbola.
  1. Standard Form of Hyperbola Equation: Identify the standard form of the equation for a hyperbola centered at the origin with a vertical transverse axis.\newlineStandard form of equation for a hyperbola with vertical transverse axis: \newline(yk)2/a2(xh)2/b2=1(y-k)^2/a^2 - (x-h)^2/b^2 = 1 where (h,k)(h, k) is the center of the hyperbola.
  2. Determining the Center: Determine the center (h,k)(h, k) of the hyperbola. Since the hyperbola is centered at the origin, we have h=0h = 0 and k=0k = 0.
  3. Finding the Semi-Major Axis: Find the value of the semi-major axis aa. The vertices are given at (0,±12)(0, \pm\sqrt{12}), so the distance from the center to a vertex is 12\sqrt{12}. Therefore, a=12a = \sqrt{12}.
  4. Finding the Semi-Minor Axis: Find the value of the semi-minor axis bb using the relationship c2=a2+b2c^2 = a^2 + b^2, where cc is the distance from the center to a focus.\newlineThe foci are given at (0,±45)(0, \pm\sqrt{45}), so c=45c = \sqrt{45}. We already know that a=12a = \sqrt{12}. We can now solve for bb.\newlinec2=a2+b2c^2 = a^2 + b^2\newline(45)2=(12)2+b2(\sqrt{45})^2 = (\sqrt{12})^2 + b^2\newline45=12+b245 = 12 + b^2\newlinec2=a2+b2c^2 = a^2 + b^200\newlinec2=a2+b2c^2 = a^2 + b^211\newlinec2=a2+b2c^2 = a^2 + b^222
  5. Writing the Equation in Standard Form: Write the equation of the hyperbola in standard form after substituting the values of hh, kk, aa, and bb. Substitute h=0h = 0, k=0k = 0, a=12a = \sqrt{12}, and b=33b = \sqrt{33} into the standard form equation (yk)2/a2(xh)2/b2=1(y-k)^2/a^2 - (x-h)^2/b^2 = 1. (y0)2/(12)2(x0)2/(33)2=1(y - 0)^2/(\sqrt{12})^2 - (x - 0)^2/(\sqrt{33})^2 = 1 kk00

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