If the eccentricity of the hyperbola frac{x^{ | vedclass

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(B) The equation of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Since $2x+3y-6=0$ is a focal chord,it must pass through the focus $(

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$\frac{24}{5}$

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(B) The equation of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Since $2x+3y-6=0$ is a focal chord,it must pass through the focus $(ae, 0)$. Substituting the coordinates of the focus into the equation of the chord: $2(ae) + 3(0) - 6 = 0$ $2ae = 6$ $ae = 3$ Given the eccentricity $e = \frac{5}{4}$,we have: $a 	imes \frac{5}{4} = 3$ $a = 3 	imes \frac{4}{5} = \frac{12}{5}$ The length of the transverse axis is $2a$. $2a = 2 	imes \frac{12}{5} = \frac{24}{5}$.

If the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $\frac{5}{4}$ and $2x+3y-6=0$ is a focal chord of the hyperbola,then the length of the transverse axis is equal to

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