Q.Find the length of transverse axis, length of conjugate axis, the eccentricity, the co-ordinates of foci, equations of directrices and the length of latus rectum of the hyperbola $\dfrac{x^2}{25} - \dfrac{y^2}{16} = 1$.

The hyperbola's focal properties mirror the ellipse's closely, but with one sign consistently flipped throughout. The latus rectum again has length $\dfrac{2b^2}{a}$, and the distance between directrices is again $\dfrac{2a}{e}$ — these two formulas are actually identical in form to the ellipse's.

The key DIFFERENCE is in the focal-distance identity: whereas an ellipse point's focal distances always SUM to the constant $2a$, a hyperbola point's focal distances always have DIFFERENCE equal to the constant $2a$ — the length of the transverse axis. This is the hyperbola's own defining "distance" property, the counterpart of the ellipse's pin-and-string construction.

Points on the hyperbola are parametrised (via the auxiliary circle $x^2+y^2=a^2$, drawn on the transverse axis as diameter) as $x=a\sec\theta,\,y=b\tan\theta$ — using secant and tangent rather than the ellipse's cosine and sine, since $\sec^2\theta-\tan^2\theta=1$ matches the hyperbola's minus sign, just as $\cos^2\theta+\sin^2\theta=1$ matched the ellipse's plus sign.

A special case worth remembering: a rectangular hyperbola has $a=b$, and its eccentricity always works out to $e=\sqrt2$, regardless of the actual size $a$.