If a hyperbola has length of its conjugate axis equal to 5 and the distance between its foci is 13, then the eccentricity of the hyperbola is:

For a standard hyperbola with its centre at the origin and transverse axis along the x-axis, we write the equation as $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1.$$ Here

• the length of the conjugate axis is $$2b,$$
• the distance of each focus from the centre is $$c,$$
• the distance between the two foci is therefore $$2c,$$
• the eccentricity is $$e=\dfrac{c}{a},$$
• and the fundamental relation among the semi-axes is $$c^{2}=a^{2}+b^{2}.$$

We are told that the conjugate axis has length 5, so

$$2b = 5 \;\;\Longrightarrow\;\; b = \frac{5}{2}.$$

The distance between the foci is given as 13, hence

$$2c = 13 \;\;\Longrightarrow\;\; c = \frac{13}{2}.$$

By definition of eccentricity we have $$c = ae,$$ so

$$ae = \frac{13}{2} \;\;\Longrightarrow\;\; a = \frac{13}{2e}.$$

Now we invoke the relation $$c^{2}=a^{2}+b^{2}.$$ Substituting $$c=ae$$ and then inserting the known values step by step we get

$$a^{2}e^{2}=a^{2}+b^{2}.$$

Rearranging gives

$$a^{2}(e^{2}-1)=b^{2}.$$

So

$$a^{2}=\frac{b^{2}}{e^{2}-1}.$$

But we already have $$a=\dfrac{13}{2e}$$ from the focus condition. Squaring this expression yields

$$a^{2}=\left(\frac{13}{2e}\right)^{2}=\frac{169}{4e^{2}}.$$

Equating the two expressions for $$a^{2}$$ and substituting $$b^{2}=\left(\dfrac{5}{2}\right)^{2}=\dfrac{25}{4},$$ we obtain

$$\frac{169}{4e^{2}}=\frac{25/4}{\,e^{2}-1\,}.$$

Multiplying both sides by $$4e^{2}(e^{2}-1)$$ eliminates the denominators:

$$169(e^{2}-1)=25e^{2}.$$

Expanding and collecting like terms gives

$$169e^{2}-169=25e^{2},$$

$$169e^{2}-25e^{2}=169,$$

$$144e^{2}=169.$$

Dividing by 144 leads to

$$e^{2}=\frac{169}{144}.$$

Taking the positive square root (because eccentricity is always >1) we get

$$e=\frac{13}{12}.$$

Hence, the correct answer is Option A.