Question:
The length of the transverse axis of the hyperbola \(3x^2 - 4y^2 = 32\) is:
The length of the transverse axis of the hyperbola \(3x^2 - 4y^2 = 32\) is:
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Always rewrite the equation of a conic in standard form before identifying parameters like \(a\), \(b\), and \(c\). For hyperbolas, the transverse axis length is \(2a\).
Updated On: Mar 24, 2026
- \(\frac332\)
- \(\frac643\)
- \(\frac8\sqrt2\sqrt3\)
- \(\frac16\sqrt2\sqrt3\)
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The Correct Option is C
Solution and Explanation
We are given the equation of the hyperbola:
\[ 3x^2 - 4y^2 = 32 \] To put it into standard form, divide through by 32:
\[ \frac3x^232 - \frac4y^232 = 1 \] Simplify each term:
\[ \fracx^2\frac323 - \fracy^28 = 1 \] The standard form of a hyperbola with a horizontal transverse axis is:
\[ \fracx^2a^2 - \fracy^2b^2 = 1 \] Comparing, we see \(a^2 = \frac323\).
Thus:
\[ a = \sqrt\frac323 = \frac\sqrt32\sqrt3 = \frac4\sqrt2\sqrt3 \] The length of the transverse axis is \(2a\):
\[ 2a = 2 \times \frac4\sqrt2\sqrt3 = \frac8\sqrt2\sqrt3 \] Therefore, the correct answer is \(\frac8\sqrt2\sqrt3\).
\[ 3x^2 - 4y^2 = 32 \] To put it into standard form, divide through by 32:
\[ \frac3x^232 - \frac4y^232 = 1 \] Simplify each term:
\[ \fracx^2\frac323 - \fracy^28 = 1 \] The standard form of a hyperbola with a horizontal transverse axis is:
\[ \fracx^2a^2 - \fracy^2b^2 = 1 \] Comparing, we see \(a^2 = \frac323\).
Thus:
\[ a = \sqrt\frac323 = \frac\sqrt32\sqrt3 = \frac4\sqrt2\sqrt3 \] The length of the transverse axis is \(2a\):
\[ 2a = 2 \times \frac4\sqrt2\sqrt3 = \frac8\sqrt2\sqrt3 \] Therefore, the correct answer is \(\frac8\sqrt2\sqrt3\).
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