Question

If the tangent at the point $(p,q)$ on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ cuts the auxiliary circle in points whose ordinates are $y_1$ and $y_2$ then show that $q$ is harmonic mean of $y_1$ and $y_2.$

Answer 1

Here we choose the tangent as $\frac{xp}{a^2}-\frac{yq}{b^2}=1$ where $\frac{p^2}{a^2}-\frac{q^2}{b^2}=1.$ ...(1)
Its intersection with $x^2+y^2=a^2$ is given by eliminating $x$ as we are concerned with ordinates
${(1+\frac{yq}{b^2})}^2\cdot \frac{a^4}{p^2}+y^2=a^2$
${(b^2+yq)}^2{a}^4+b^4{y}^2{p}^2=a^2{p}^2{b}^4$
$y^2(a^4{q}^2+b^4{p}^2)+2yq{b}^2{a}^4+a^4{b}^4$
$-a^2{p}^2{b}^4=0$ .....(2)

Above is a quadratic in $y$.

We have to prove that $q$ is H.M. '$H$' of $y_1$ and $y_2$
Now

$H=\frac{2y_1{y}_2}{y_1+y_2}=\frac{(a^4{b}^4-a^2{p}^2{b}^4)}{-2q{b}^2{a}^4}$ by (2)

$=\frac{a^4{b}^4(1-\frac{p^2}{a^2})}{-2q{b}^2\cdot a^4}=\frac{b^2}{-q}[-\frac{q^2}{b^2}]=q,$ by (1).

$\therefore$ $q$ is H.M. of $y_1$ and $y_2.$