Q.Find the equation of the hyperbola referred to its principal axes whose vertices are $(\pm7,0)$ and end points of conjugate axis are $(0, \pm3)$.

Applying the focus-directrix construction with eccentricity $e>1$ produces the standard equation of a hyperbola:

The derivation runs in close parallel to the ellipse's, but the key algebraic difference is that $1-e^2$ is now NEGATIVE (since $e>1$), which flips a sign partway through and produces $b^2=a^2(e^2-1)$ — the mirror image of the ellipse's $b^2=a^2(1-e^2)$.

Structurally, the hyperbola differs from the ellipse in an important way: it meets the $X$-axis at its two vertices $(\pm a,0)$ but never meets the $Y$-axis at all (setting $x=0$ gives no real solution for $y$). Because slicing a double cone with a plane parallel to the axis cuts through BOTH nappes, the hyperbola naturally has two separate open branches rather than the ellipse's single closed curve. The segment joining the vertices (length $2a$) is the transverse axis; the perpendicular segment through the centre (length $2b$) is the conjugate axis — together the principal axes, …