Prove that the part of the tangent at any point of the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1` intercepted between the point of contact and the transvers axis is a harmonic mean between the lengths of the perpendiculars drawn from the foci on the normal at the same point.

Explanation

Equation of tangent to hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` at point `P(a sec theta, b tan theta)` is
`(x sec theta)/(a)-(y tan theta)/(b)=1" (1)"`
Equation of normal at point P is
`(ax)/(sec theta)+(by)/(tan theta)=a^(2)e^(2)`
Tangent meets x-axis at Q `(a cos theta, 0)` and normal meets x-axis at `G(e^(2)a sec theta, 0)`.

In the figure, PQ = p and length of perpendiculars from foci `F_(1) and F_(2)` on the normal are `p_(1) and p_(2)`, respectively.
Now, `(p)/(p_(1)=(QG)/(F_(1)G)`
`=(e^(2)a sectheta-a costheta)/(e^(2)a sec theta-ae)`
`=(e^(2)-cos^(2)theta)/(e^(2)-e costheta)=(e+costheta)/(e)`
`therefore" "(p)/(p_(1))=1+(costheta)/(e)" (2)"`
Similarly, we get
`(p)/(p_(2))=1-(cos theta)/(e)" (3)"`
Adding (2) and (3), we get
`therefore" "(p)/(p_(1))+(p)/(p_(2))=2`
`rArr" "(2)/(p)=(1)/(p_(1))+(1)/(p_(2))`
Thus, `p_(1),p and p_(2)` are in H.P.